AIDS TO THE i Maruenatics or HYGIENE iE R. B. FERGUSON _ FIFTH EDITION i) ss BAILLIBRE, TINDALL & cox JU MINIM UT Med K21821 aw F tt a Dali es AIDS TO THE MATHEMATICS OF HYGIENE AIDS THE MATHEMATICS OF HYGIENE BY R. BRUCE FERGUSONDICA/ Atm, M.A., M.D., B.C. (Cantas.), D)P.H. (ENG.), ETC. LED FIFTH EDITION LONDON BAILEIE RE TINDALL & COX 8, HENRIETTA STREET, COVENT GARDEN 1919 Ha dike Aa on lees f WELLCOME INSTITUTE LIBRARY colt welMOmec | PREFACE TO, THE Fl TH EE DEPION IN preparing the fifth edition of this book, I have taken the opportunity not only to revise it, but to bring it up to date by the introduction, amongst other things, of a note on the new method of recording barometrical obser- vations, and the estimation of the calorific values of foods—a matter which is attracting a good deal of attention at the present time. | Many of the existing examples have been replaced by others, of a similar form, taken from papers recently set for the Diploma in Public Health by different examining bodies, in the hope they may prove a useful guide to intending candidates as to the nature and extent of the mathematical knowledge required-from them. R. B. F. FoREST SIDE, EPPING. October, 1919. =. PREFACE TO -LHE SECOND EDITION THE issue of a second edition of this little book has afforded me the opportunity of revising it throughout. A certain amount of new matter and some additional examples have been introduced, whilst the chapter on Vital Statistics has been re-written and enlarged. On the other hand, such portions have been deleted as appeared to stray beyond the limits I originally laid down for myself—viz., to deal with the study of Hygiene solely from the szathematical side of the subject. This side has always seemed to me to be treated rather inadequately by even the standard text-books. In these, a formula is inserted in its entirety, without a hint as to how such a result has been arrived at, or by what method it was compiled. Without this knowledge, the only alter- native is to commit it to memory—with the risk of forget- ting it—before entering for an examination on the subject. It is also generally taken for granted that the reader is well acquainted with all the methods of chemical and physical calculations; but since these subjects are amongst the earliest of one’s scientific studies, it is quite within the range of possibility that these details may have passed somewhat from recollection ; so that, should he be pre- paring for examination in Public Health, they must be all Vil vill PREFACE TO THE SECOND EDITION sought out at infinite trouble from, possibly, a dozen different sources (chemistry, physics, algebra, statics, dynamics, trigonometry, etc.). It has been my endeavour to collect these formule, and to put them together in these pages, chiefly in the form of examples. I have also attempted to show, in as simple a manner as possible, how they may be applied and made use of, in the solution of the various problems so frequently met with during the course of one’s study. The book is not meant to be a guide to Hygiene, nor yet to embrace every branch of the subject. It is only what its title claims for it—viz.,a guide to those portions of Public Health which require mathematical treatment. As such, it is gratifying to find that it has already been of some assistance; and it is hoped that it will continue to prove useful, especially to those who have not the time, nor yet, perhaps, the inclination, to commence their mathematical studies afresh. The calculations connected with the practical examina- tion of milk, water, etc., have been omitted, since these are so inseparably connected with the actual laboratory work, that their inclusion here would necessitate the in- troduction of the whole methods of analysis—a subject already fully dealt with in the many special text-books. hes BU October, 1903. CONTENTS CHAPTER |} LAWS OF GASES Definitions — Gases — Vapours — The Relationship between their Volume, Weight, Pressure, and Temperature—Diffusion - - - - - CHAP TEAR I HYGROMETRY Dry Air—Aqueous Vapour—Moist Air (saturated and non-saturated)— Dew-point—Humidity and Dry- ing-power of Air—The Rain-gauge—The Barometer and Thermometer—Conversion of Thermometer and Barometer Scales—The C. G. S. System — Glycerine Barometer — Water Barometer — Alcohol Barometer—Correction of Barometric Readings for Altitude and Temperature—Method of Ascertaining Altitude is the Barometer—The Vernier - : - - - - - CHAPTER III HYDROSTATICS AND HEAT Speciac Heat—Latent Heat— Specific Gravity— Hydraulic Press” - . : “ é = : ix PAGE We x CONTENTS CHARTER INV VENTILATION Velocity and Delivery of Air—Size of Inlets—Amount of Carbonic Acid Gas expired—Quantity of Fresh Air required—Amount of Impurity present— Impurity produced by Artificial Illumination— Combustion of Oil—Montgolfier’s Formula—Fric- tion—Ventilation by Propulsion—Fans - - GHAP TER Vv. RAINFALL AND SEWERAGE Rainfall and Water-supply—Storage—Flow in Sewers —Hydraulic Mean Depth—Velocity, Gradient, and Size of Sewers, and the Relationship between them -—— Excreta — Estimation of Nitrogen and Ammonia in Excreta- - - - : - - CHAP LRN: ENERGY, EXERCISE, AND DIET Energy expended in Walking—An ‘Ordinary Day's Work ’—Calculation of Diets—Energy available from Food—The Fuel-value of Food— A Calorie - CHAPTER VII THE CONSTRUCTION OF A HOSPITAL WARD Length— Width—Height—Number of Patients—Floor- space—Cubic Space — Windows — Beds — Fever- ward - - - - - - - - 44 67 84 96 CONTENTS CHAPTER -ViUI ALCOHOL AND PROOF-SPIRIT Percentage Composition of Proof-spirit by Volume and by Weight — Overproof — Underproof — Adultera- Hoty Sa ae ee eye i ag one CHAPTER ix CHEMICAL CALCULATIONS Disinfection by Sulphur—Percentage Compositions— Standard Solutions—Solutions for Testing for Chlorine, Carbonic Acid Gas, and for ‘ Nessler- ising "Percentage Strength of Solutions—Con- version of ‘grains per gallon’ into ‘parts per 100,000’—Normal Solutions - - - - . CHAPTER X LOGARITHMS Definition—Their Fundamental Properties and Practi- cal Application—Method of Using a Logarithmic Table - z : ‘ : < a . CHAPTER xh VITAL STATISTICS Increase in Population—Methods of Estimating a Popu- lation—Mean Annual Population for a Period of Years — Birth-rates and Death-rates — Zymotic Death-rate — Infantile Mortality Rate — Mean Annual Rates—Annual Rates for Short Periods— Death-rate of a Combined District — Weighted Averages — Corrected Death-rates — Standard Death-rate—A Standard Million --Comparative Mortality Figure (general and occupational)— Poisson’s Rule—Averages and Probable Error - % X1 PAGE 102 107 118 128 xii CONTENTS CHAT TEREX LIFE TABLES Probability—Construction ofa Life-table by the ‘short ’ method—Data required—Preliminary Calculations —pxColumn—/, Column—d,Column—P, Column —Q, Column—Ex Column—Summary—Curtate Expectation of Life—-Probable Life-time—Prob- abilities of Life as deduced from the Life-table - CHAPTE RK AIM MENSURATION Circle—Ellipse—Square—Rectangle—Parallelogram— Triangle — Sphere — Cylinder — Cone — Dome — Microscope-field — - - - - - - : Weights and Measures - - . - = ° . Atmospheric Pressure and Boiling-point - - - Table of Vapour Tensions” - - - - - am PAGE 159 ATS TO THE MATHEMATICS OF HYGIE Si: CHAPTER I LAWS OF GASES Definitions : Gases are bodies whose molecules are in a constant state of motion, in virtue of which they possess the most perfect mobility, and are continually tending to occupy a greater space. They are also called ‘elastic fluids.’ A minute quantity of a gas may be made to occupy any space, how- ever large, by sufficiently reducing the pressure to which it is exposed. Vapours are the aeriform state into which liquids—such as water or alcohol—are converted by the application of heat. A gas obeys the same laws—as to temperature and pressure—as atmosphericair. A vapour only obeys these laws within certain limits. Non-saturated vapours re- semble gases, and therefore obey the laws of Boyle and Charles (p. 5). Saturated vapours do not, for if the pressure be too great or the temperature too low, a portion J 2) AIDS TO THE MATHEMATICS Gla YGIENE of the vapour will at once pass into the liquid state ; whilst if the pressure be reduced or the temperature raised, they will no longer be saturated. Thus for a given temperature, vapours, when saturated and in contact with their liquid, can neither be compressed nor expanded. Gases and the Relationship of their Volume, Weight, Pressure, and Temperature. Volume and Weight: The weight of a given volume of any gas or vapour can readily be calculated if we know— (i.) The weight of a given volume of some gas which can be taken as a standard ; (ii.) The relative density of the gas, compared with this standard. In Chemistry it is usual to take, as the standard, the weight of a given volume of “ydrogen at a certain tem- perature and pressure ; and the weight of any other gas, at the same temperature and pressure, is obtained by multiplying this by the relative density of the gas whose weight is being ascertained. The standard always adopted is that known as the Gritty ealicalss: I litre of hydrogen at o° C. and 760 mm. weighs 008958 gramme. Now, the densities of all elements in the gaseous state are identical with their atomic Peleus ; and since OQ=16, and S= 32, therefore I litre of oxygen at o° C. and 760 mm. weighs 16 x 008958 = 1°43 grammes ; and 1 litre of sulphur vapour at 0° C. and 760 mm. weighs 32 X 0'08958 = 2°86 grammes. LAWS OF GASES 3 The density of a compound gas is one-half its mole- cular weight ;-and since CO,=44, and H,O=18, there- fore the density of CO,=22, and the density of water- vapour (steam)=9, compared with that of H as unity. .. I litre of CO, at o° C. and 760 mm. weighs 22 X0°08958=1'97 grammes. For convenience of calculation, the relationship between the weight and volume of a gas may be expressed as follows : 1 litre of H at o° C. and 760 mm. weighs 0'08958 grm. ; I litre weighs I gramme ; ** 0'08958 Z.é., 11°2 litres of H weigh I gramme. 2 Sele r= , O » 16 grammes ; and 162°. ,, tS) a) 38 % iS ee COs 3... 22 2 eae steam’, - 9 5 That is, the weight of 11°2 litres of any gas at 0° C, and 760 mut. ts tts density expressed in grammes. In Physics, it is usual to take azy as the standard, and the density or weight of a gas or vapour is the relation between the weight of a given volume of this gas or vapour, and that of the same volume of air at the same temperature and pressure. The standard here adopted is: I cubic foot of dry air at 32° F. and 760 mm. weighs 566°85 grains. Since air is 14°47 times heavier than hydrogen, the weight of any gaseous element, or compound, may be found as follows : 4 AIDS TO THE MATHEMATICS OF HYGIENE Density of oxygen: density of air : : 16: 14°47; Density of water-vapour : density of air: : 9 : 14°47; and so for any other gas. [Note on PRroporTION.—Four quantities are said to be proportional when the ratio of the first to the second is equal to the ratio of the third to the fourth. Thus, if a, b, c, and d@ are proportional — that is, a:b::¢:d—then =a or b one or ad=bc. | cad Example: Find the weight of 1 cubic foot of SO, at 32° F. and 760 mm. weight of SO, 32_—;, weight of air 14°47" pe iscubic foot SOsn 32 5 50635 14°47” Since . I cubic foot SO,= 566°85 x See. = 1253°57 grains. 14°47 Since the coefficient of expansion of all gases is the same as that of air, it does not signify at what tem- perature or pressure the relative densities are compared, provided they are both at the same temperature and pressure ; ¢.g., the relative densities of air and aqueous vapour (the pressure being constant) would be the same at, 60° \F,.as) at 32-7) ., provided: both aresateGo wk. sor 32° F. respectively; but the same fraction would not represent the relative density between air at 32° F. and aqueous vapour at 60° F. Example : The relative density of air, where hydrogen is the standard, being 14°47, find the relative densities of oxy- LAWS OF GASES 5 gen, hydrogen, carbon-dioxide, and aqueous vapour, taking air as the standard.* Density of oxygen : density of air :: 16: 14°47 ; be density of O es 16 > density of air 14°47 Take density of air=1, then density of O= pes) =I'l. 14°47 Density of hydrogen : density of air :: I : 14°73 ». density of H=—!— =0°069. z 14°47 Density of carbon-dioxide : density of air : : 22: 14°47; : 22 *, density of COg=——— = 152. y 2 a7 5 Density of aqueous vapour : density of air :: 9 : 14°47; *, density of aqueous vapour — 9_ = 0°622. 14°47 (In the following examples, W, V, P, and T represent the present—or existing—weight, volume, pressure, and tem- perature respectively ; while w,v, 2, and ¢, represent the new weight, volume, etc., which are to be ascertained.) ~ Example : What are the laws of the volume of gases—(a) to pressure, (4) to temperature? Illustrate by an example of each. + (z) Volume and Pressure : 7 Boyles (or Mariotte’s) law states that the temperature being constant, the volume (V) varies inversely as the pressure (P). That-is; Vi: Woe ren ret re * D.P.H, Exam., Cambridge. ft D.P.H, Exam. (Roy. Coll. Phys. Surg,), 6 AIDS TO THE MATHEMATICS OF HYGIENE Example: A gas measures 100 c.c. at 730mm. Find its volume at 760 mm. VEIT) Saree al OO Eee ie UO 408 a 00 730) da rere 96:C.C; (6) Volume and Temperature : Charles's (or Gay Lussac’s) law states that all gases I : expand ae part of their volume at 0° C. for every increase I in temperature of 1° C., or fer part of their volume at 32° F. for every increase in temperature of 1° F.; | fe) 22.5.1 VOl. atOa.G.. becomes ateiO =C.al = sc Or 273+10 273 60 — 32 491 +(60 - 32) , or 49! 491 volsi, and’ 1) vol: at 3220 h.pbecomessat 00. FE. ict vols. Charles's law may therefore be stated as follows: The pressure being constant, the volume of a gas varies directly as the absolute temperature (the absolute tem- perature for the Centigrade scale being the tempera- ture + 273)—2.¢., Vigra 9253027,.3 te eee tea, or for the Fahrenheit scale : Viu:: 491+(T- 32) : 491+ (¢—-32). Examples : (i.) What volume would 1,000 c.c. of a gas at o° C. expand to at 80° C.? : LAWS OF GASES 7 Misete? 27esi ss 2732-8. £,.1,000 2 @ 72 273-+0 3.2734 80; > + 273 + 3535 __ 1,000 X 353 _ @e med 273 1,293 C.G. (ii.) What volume would 1,000 c.c. of a gas at 4o° C. expand to at 120° C. ?* The rise in temperature is 80° C., as in the preceding example ; but in this case the expansion takes place in the ratio of 273+ 40 : 273+ 120. Vi 27s a2 re o. 1,000): @ : 2 2734-40 + 2734 120 > + 313 + 393; pogo sees 393 _ 1955 Cc: 313 Pressure and Weight: It has been seen from Boyle’s law that, the temperature remaining the same, the volume of a gas varies inversely I as the pressure, or P « V3 and as the quantity of gas remains the same, its density must obviously increase as its volume diminishes—that is, density varies inversely as volume, or De re It follows, therefore, that density varies directly as pressure, (or Dx P) that is to say, that for the same temperature, the density of a gas, and there- fore its weight, is proportional to its pressure ; or, Ws wes: Pos pg: * D.P.H. Exam. (Roy. Coll, Phys. Surg.). 8 AlDS TO THE MATHEMATICS OF HYGIENE Temperature and Weight: Since the volume varies directly as the absolute tem- perature—Charles’s law—and the density diminishes in proportion to the increase of volume, therefore the density (and, consequently, the weight) varies inversely as the absolute temperature ; or, Wii eU ti rep gt be. 273 ee or for the Fahrenheit scale: W : wii: 491+ (¢-—32) : 491+ (T—32). Pressure and Temperature : When the volume is constant, the pressure varies directly as the absolute temperature ; or, Pty 0 Musk Oil e ea ce Le Example : Steam at 100° C.—and, therefore, at a pressure of 760 mm.—is removed from contact with water, and passed into the closed chamber of a Disinfector, and heated until the pressure-gauge registers 790 mm. Find the temperature of the steam in the chamber. The steam, being superheated, is consequently a non- saturated vapour, and as such obeys the laws of gases 30ah)s POPE 127341 ona se ¢°2, 1 FOO! 2°790.4)3. 2734+ 100 9273-7 « “"s 700 (273 +2) =790 X 373 5 si IOS Io LAWS OF GASES 9 Example (involving variations in temperature, pres- sure, and volume) : A gas at 7° C. and 760 mm. occupies 15 litres. Find its volume at —8° C. and 720 mm. 1 (i.) Find volume'due to change of pressure ; Vv fy pee 3 a tS fy: '920°.-760 § a, Pt Aaa ra os = 6 = 15°83 litres. (ii.) Find volumé of this new volume, due to change of tempergture : VED 32 27341 27548: os F583 Oe: 273427 373-33 23. 260.205 3 - = 14°98 litres Examples of this nature may also be solved by the algebraical process known as Variation. It has been seen that where the pressure (P) is constant, the volume (V) varies as the absolute temperature (A), or VocA; y sai ane die : that is, the ratio q 18 constant. If #z=this constant ratio, then hem, and V=mA. Similarly, where the temperature is constant, V varies inversely as P, or V « ; “that 18, Vo = 7 > oD VP =m. Where both temperature and pressure vary, V will vary A 1 as the product Axz; that is Vo > and V=m p a / - Cd Therefore, 7z= to AIDS TO THE MATHEMATICS OF HYGIENE In the preceding example— V=15, P=760, A=273+7=280; __15 xX 760 AUD 280 = 407. A New volume =72 —. Pr Where m= 40°7, A=273-—8=265, P=720; Mexp ADT R205 |, ko 7 $i Visaea cee) = 14°98 litres. DIFFUSION OF GASES. Example : What is the law relating to the diffusion of gases ?* If two different gases be separated by a porous diaphragm, an exchange takes place between them, and ultimately the composition of the gas on both sides of the diaphragm is the same ; but the rapidity with which this diffusion occurs varies with the gases. Graham’s law states that the force of diffusion varies inversely as the square roots of the densities of the yases. Thus H and O will diffuse as follows ; H:0O:: 16: A1; se hpped inc nT: Or, for every one part of O which has passed into the H, four parts of H have passed into the O. * D.P.H, Exam. (Roy. Coll, Phys. Surg.). CHAPTER II HYGROMETRY DRY AIR. To find the Weight of a Given Volume of Dry Air at a Given Temperature and Pressure. From what has been shown in the previous chapter, a general formula may now be constructed which will embrace all the variations in volume, pressure, and tem- perature to which any weight of gas may be subjected, as follows : 1 cubic foot of dry air at 0° C. and 760 mm. weighs 566°85 grains; and since weight varies directly as volume, 2. wow Some WS OF) UE ESOOTOR 2 ee as oa and since weight varies directly as pressure, 2 2 WV sepa; OF WEG OO SS OS 760.6%. > Ga.)3 and since weight varies inversely as absolute tempera- ture, ees WV 22273+T : 27342; or, w: 56685 :: 273+0: 27342; 37 BL o.o ecca: wi) II 12 AIDS TO THE MATHEMATICS OF HYGIENE If these three proportions—(i.), (il.), and (iil1.)—be combined, we have: Ws GOG86 wie 2k si 2 8700.5 ihe 7 ees Hp Aeon cabot use) alge Mii psy esr Tx OEE ae Bi sift ERIS aay Seren gt or tor the Fahrenheit scale: PREC EOS EL a mete on ae ESET AUS pcan ae ESO SY (where z=weight in grains, and v=vol. in cubic feet). Example : Find the weight of 1 cubic foot of dry air at 60° F. and 735 mm. Lyag 735 AO Le senes i w=566'85 x1 Oe mason grains. — — = AQUEOUS VAPOUR. To find the Weight of a Given Volume of Aqueous Vapour at a Given Temperature and Pressure. To find the weight of a vapour, the weight of the same volume of dry air at the same temperature and pressure must be sought, and this is then to be multi- plied by the relative density of the vapour. Example : Find the weight of a cubic foot of aqueous vapour at bok. From a table of vapour-tensions (p. 185), it is found that the maximum pressure which aqueous vapour can excrt at 60° F.=13'167 mm. HYGROMETRY 13 Therefore, the weight of a cubic foot of dry air at 60° F, and 13167 mm. must first be found, as follows (ps LL) 566°85 x 13°167 X 491 ~ 760 x [491 + (60— 32)] then this result must be multiplied by the relative density of aqueous vapour compared te air — Viz, by 0°622 (p.5)- Therefore, 1 cubic foot of aqueous vapour at 60° F. weighs 9°29 xX 0°622= 5°77 grains. Example : What weight of aqueous vapour is contained in a cubic foot of air which is saturated at a temperature of Goo. ? Dalton’s law states that ‘the tension, and conse- quently the quantity, of vapour which saturates a given space, are the same for the same temperature, whether this space contains a gas or is a vacuum.’ Thus, the _ formation of vapour does not depend upon the presence of air, or upon its density. If water be introduced into two similar vessels, the one containing air and the other a vacuum, the quantity of vapour formed will be the same in each case. Therefore, the weight of aqueous vapour in a cubic foot of air is the same as if the space had been empty of air; so the question resolves itself into, What is the weight of a cubic foot of aqueous vapour at 60° F, ? [Ans.: 5°77 grains. ] —9'29 grains ; SATURATED AIR. To find the Weight of a Given Volume of Saturated Air at a Given Temperature and Pressure. The mass of air may be divided into two parts—viz., a volume of dry air, and a volume of aqueous vapour; and 14 AIDS TO THE MATHEMATICS OF HYGIENE the sum of the weights of these two volumes is the weight required—z.e., the actual pressure of the mixture is the sum of the pressures due to the gas and vapour considered separately. Let P=the pressure of the moist air ; and #=the elastic force of the vapour which saturates it. Then the azz alone in the mixture only supports a pressure of P —Z. Example : Find the weight of 1 cubic foot of saturated air, at 60° F. and ordinary atmospheric pressure. (i.) Find the weight of 1 cubic foot of dry air at 60° F. and pressure P —Z (p. 11). P=760, and J=13'167 (p. 185). Ah ee ee an W700 RT 491 eel == SOU Dax ne Saran Goesas) = 526°07 grains. (ii.) Find the weight of 1 cubic foot of aqueous vapour at 60° F. and pressure Z (p. 12). __ 566°85 x 13°167 X 491 *~ 760 x [491+ (60— 32)] .. Weight of 1 cubic foot of saturated air at 60° F. and 760 mm. is: x 0:622=25°77 tains. 526°97+ 5°77 =532'74 grains. JVote.—It is seen that 1 cubic foot of saturated air at 60° F. and ordinary atmospheric pressure weighs 532°74 grains, whereas 1 cubic foot of dy air at 60° F. and 566°85 x 491 491 +(60—32) the same pressure weighs = 53627 grains. That is to say, the saturated air weighs less than an equal volume of dry air. The explanation of this is as follows ; HYGROMETAY 15 Dry air expands on taking up moisture, and when 1 cubic foot of dry air at 60° F., weighing 536°27 grains, takes up I cubic foot of aqueous vapour (at the same temperature and pressure) weighing 5°77 grains, the weight of the resulting moist air will be 536°27+5°77= 54204 grains, but the volume of the mixture will be, not 1 cubic foot, but somewhat more than I cubic foot. The moist air is under a pressure of 760 mm., and it has been already seen that the elastic force of the vapour which saturates the air is 131167 mm. Therefore, the air in the mixture supports a pressure of 760 — 13°167=746'833 mm. only, and, as volume (v) is inversely as pressure, then @: t cubic foot : : 760°: 746833 : Ae v= ea 1°0176 cubic feet. That is, the resulting mixture of the dry air and the vapour produces 1'0176 cubic feet of saturated air, weighing 542°04 grains, and 1 cubic foot of this same moist air would only weigh 542°04 : Be grains. To find the Dew-point. For any given temperature, air will only hold a certain quantity of aqueous vapour ; the higher the temperature of the air, the greater will be the amount of vapour which it can contain; when the air contains its greatest possible amount, it is said to be ‘saturated’ and the temperature at which saturation occurs is called the ‘dew- point.’ The dew-point may be obtained directly by means of a Hygrometer (Daniell’s, Regnault’s, or Dine’s), or in- directly by the dry-and-wet bulb Hygrometer. 16 AIDS TO THE MATHEMATICS OF HYGIENE In the case of the former, the dew-point is the tem- perature at which the thermometer in the blackened bulb stands, at the moment when deposition of moisture on the bulb occurs. Whereas, in the case of the dry-and- wet-bulb, the wet-bulb does zo¢ indicate the dew-point, which, therefore, cannot be ascertained by simple in- spection of the thermometer. Ifthe air be saturated no evaporation is possible, and the two thermometers will read alike ; if not saturated, the wet-bulb will read lower than the dry-bulb, but not so low as the dew-point ; in fact, the temperature of the wet-bulb is always above the dew-point. When the dry bulb stands at 53° F. the dew-point is as much below the wet-bulb as the wet-bulb is below the dry-bulb. Above this temperature, the wet- bulb approaches nearer the dew-point, and the reverse is - the case below that temperature. Glaisher has empirically compiled some tables, whereby the difference between the dew-point and the wet-bulb bears a constant ratio to the difference between the wet- bulb and the dry-bulb; so that, if the reading of the dry-bulb be given, the dew-point can be calculated. According to him, the temperature of the dew-point is obtained by multiplying the difference between the wet- and dry-bulb temperatures bya constant factor (‘Glaisher’s factor’), and subtracting the product thus obtained from the dry-bulb temperature, thus: Dew-point=Ta —[(Ta — Tw) x F]. Where Tq =temperature of dry-bulb (Fahrenheit). T,, =temperature of wet-bulb . F =factor (found opposite the dry-bulb tem- perature in the table). lf T) =60°°R. and T,=s4° F., then fronttable) F=1°88 and dew-point = 60 —[(60 — 54) x 1°88]=48°72° F, HYGROMETRY 17 From this formula Glaisher calculated his tables, which give the dew-point on inspection. The dew-point may also be found by Apjohkn's formula. For this purpose a table of vapour-tensions (p. 185) is required. His formula is as follows : F=f- & « =) for temperatures above 32° F., and h is F=/-(% x | for temperatures below 32° F. Where F =tension of vapour at dew-point, f=tension of vapour at temperature of wet- bulb. d= difference (Fahrenheit) between the wet- and-dry-bulb thermometers. i =height of barometer (in inches). Near the sea-level, the fraction - differs very little from unity, and may be neglected; so the formula may be simplified thus : a L sae or Fafa oo The constant 87 (or 96) represents the specific heat of air and vapour. Having found F, the table of vapour-tensions must be again referred to, and the temperature opposite the tension F will be the dew-point. To recapitulate the more important, points— The dew-point may be found in two different ways: (i.) By direct observation of thermometer, in Daniell’s, Regnault’s, or Dine’s Hygrometer. eu] 18 AIDS TO THE MATHEMATICS OF HYGIENE (ii.) Indirectly, by the dry-and-wet-bulb Hygro- meter— (z) By means of Glaisher’s tables. (6) By means of Apjohn’s formula. It must be noted that Glaisher’s formula gives the dew-point directly, no table of vapour-tensions being required ; whereas Apjohn’s formula requires the use of a vapour-tension table, the result being zo¢ the dew-point, but the vapour-tension at dew-point, from which the dew- point can be ascertained. And attention is again drawn to the fact, that the wet-bulb thermometer does not give the dew-point (except when the air is saturated), and therefore the wet-and-dry-bulb Hygrometer must not be confounded with-any of the a@rect Hygrometers—e.g., Daniell’s, where the thermometer does give the dew- point. | To find the Humidity and the Drying Power of the Air. Example: The temperature of a room is 60° F., and the dew- point is 50° F. Find the degree of humidity of the room, and the drying power of the air at the time of observation. weight of water actually present in given vol. of air Kine = ae weight of water which would saturate the same vol. The water-vapour actually present is enough to saturate the air at 50° F., but not enough to do so at 60° F. This gives the actual tension of the water-vapour present in the air at 60° F. ; for it must be the same as the maxcmusm tension at 50° F. In other words, the actual vapour pressure in any portion of air is equal to the maximum vapour pressure at dew-point. HYGROMETRY | 19 Let Ps>=maximum tension at 50° F., and Pa= 99 99 99 60° F, Then, the weight of vapour actually present’ (e.g, 11 a cubic foot) will be (p. 12): 566°85 xX Psp X 491 760 x [491 + (60—32)] X 0'622 = Psp X 0°439 grains, and the weight of vapour which would saturate the same volume will be : 566°85 X Pe X 491 760 x [491 + (60— 32) Ne as Poy X 0°439 _ Ps) . maximum tension at 50° F. Peo X0'439 Pe) maximum tension at 60° F. -Maximum tension at dew- point ™ maximum tension at existing temperature’ x 0°622 = Pep X 0'439 grains ; which, from the table (p. 185), will be found to be = 696 or 69°6 (approximately 70) per cent. So the relative humidity of the air can always be found from a table of maximum vapour-tensions, if only the dew-point and the temperature of the air at the time of observation be known. The drying power of the air means the additional weight of vapour necessary to cause saturation. If relative humidity=70 per cent., then amount of vapour actually present in a given volume of air is 70 per cent. of saturation, and the difference (or 30 per cent.) will represent the drying power. Thus, if W=weight of vapour required to saturate a cubic,.foot of air at 60° F., and relative humidity =70 per cent., then the amount of vapour actually present 20 AIDS TO THE MATHEMATICS OF HYGIENE will be — and the drying power of the same air will O : : ; be ca And since the weight of vapour in a saturated cubic foot of air at 60° F. is 5°77 grains (p. 13), pairs the amount of vapour actually present is 5°77 x o. = 4'03 grains, and the drying power is the difference between these—viz., 5°77 — 4°03 =I'7 grains. To find the Weight of a Given Volume of Moist (non-saturated) Air at a Given Temperature and Pressure, the Hygrometric State or Relative Humidity being given. Example: Find the weight of 1 cubic foot of moist air at 60° F. and ordinary atmospheric pressure, the relative humidity of the air being 60 per cent. It has been seen (p. 14) that 1 cubic foot of moist air at 60° F. is nothing more than a mixture of (i.) 1 cubic foot of dry air at 60° F., under the existing barometric pres- sure #i2us the tension of the vapour present, and (ii.) I cubic foot of aqueous vapour at 60° F., the tension of which must be found from the hygrometric state, or relative humidity. It has also been seen (p. 19) that: max. tension at dew-point Relative humidity =—— se OU max. tension at existing temp. * The existing temperature in this example is 60° F., and HYGROMETRY 21 from the table (p. 185) it is found that the maximum tension of vapour at 60° F.=13°167 mm. max. tension at dew-point , 13'167 *, max. tension at dew-point=rel. hum. x 13°167, and relative humidity = 60 per cent. ; 60 X 13°167 100 *, Relative humidity = ‘. max. tension at dew-point = =7'9 mm. Now, the actual vapour-pressure in any portion of air ; is equal to the maximum vapour-pressure at dew-point., Therefore : Actual vapour-pressure in air under observation =7°9 mm. The question, then, resolves itself into: (i.) Find weight of 1 cubic foot of dry air at 60° F. and 760—7°9=752°I mm. (p. Ir). pp REO Ee ee i) = 566°85 X 566 eras ay = 530°69 grains. (ii.) Find weight of 1 cubic foot of aqueous vapour at 60° F. and 7’9 mm. nu (p. 12). 491 w=566°85 x 760" ja ee aay ee 3°46 grains, Therefore, the weight of 1 cubic foot of air (containing 60 per cent. of moisture) at 60° F.= 5 30°69 + 3°45= 53415 grains. Incidentally, it may be remarked that the tempera- ture at which the aqueous vapour would exert a maximum pressure of 7°9 mm. will be the dew-point, which from the table (p. 185) will be seen to be about 46° F. 22 AIDS TO THE MATHEMATICS OF HYGIENE To Graduate a Rain-Gauge. Carefully measure the area of the top of the gauge or receiving surface. Suppose this to be’50 square inches. If, now, this area be covered with water to the height of I inch, the quantity of water will be 50 cubic inches. So, to graduate the gauge, 50 cubic inches of water must be put into the glass measure, and a mark placed at the level of the top of the fluid; this mark will represent 1 inch rainfall. The space below may be divided into numerous equal parts, each representing equal fractions of an inch. The 50 cubic inches of water may be obtained by measurement, as follows : At 4° C. (or 39°2° F.), or the maximum density point of water, 1,000 fluid ounces=1 cubic foot=1,728 cubic inches ; 1,000 X 50 .*. 50 Cubic inches = as = 28'9 fluid ounces. ? If the area of the receiving surface had been 100 square inches instead of 50, then 100 cubic inches of water would represent 1 inch of rainfall. The receiving surface may thus be made of any size; or, on the other hand, any quantity of water may be made to represent any fraction of an inch—e.g., what should be the diameter of the receiving surface so that 1 fluid ounce represents 4 inch? Let d=diameter in inches, o na? Then area of receiving surface= a (p. 178), and quantity of water standing 4 inch high on receiving ma? 1 wa ne ue surface will be a Xo= 5 cubic inches, and this 1s equal to 1 fluid ounce. AYGROMETRY oo. 2 me a na? o. SS el fluid ounce = 1°728 cubic inches ; 2) » 72 32% 1728 232% 0728 ean SS - Ear ate sei . d= s/17°6=4'19 inches, THE BAROMETER AND THE THERMOMETER. Conversion of Thermometer Scales. hreezing-point=32° F.,.0° C.,-and o° R. Boiling-point= 212° F., 100° C., and 80° R. Therefore: F-32 °C :R 2: 212—32¢ too: 80: 180 : 100: ao; Qe gra ae i heretore.:: ! Conversion of the Reading of One Scale into that of Another. fahrenhett—Centigrade : F-32_C, . 9 and C= 2 oe Fi 2 eee t— Réeaumur: F - 32 mS = , ae" QA 3€ = “. 4 * . 24 AIDS TO THE MATHEMATICS OF HYGIENE 0“. Fa2R+ 32, and Rao (F -32). Centigrade—Réaumur : Lxamples : 1. Find the equivalent of 98°4° F. on the other scales ca 5 MF 32) _5 (98'4 ~32)_5 x 664 _ 5a : aoe fede) BAGS eg 2) od Pd Drs 9 9 9 a GGA. e = 36°07 C205" Re 2. I'ind the equivalent of 10° C. on the other scales P= C+ 32529224 32=18+32= 503 Rae T0=-6* 5 2 iO, = con hea oo ks 3. At what temperature is the number on the Centi- grade and Fahrenheit thermometers the same? Let x=the temperature on F. scale, 5 (% 3°32) then co =temperature on C. scale ; and since the temperatures are the same, > a] ies beouk 2) 9 gat HYGROMETRY 25 9x =5x—160 9% —5x= —160 4x = — 160 r= —40°, (It may be noted that this temperature is that of the freezing-point of mercury.) 4. What is the temperature when the number of degrees on the Centigrade scale is as much below zero as that on Fahrenheit’s is above zero? Let F =temperature on F scale, C= a » C-scale. And since the temperatures are the same, but on oppo- site sides of zero— ee Cina 5 (i.) and C =2 G32) th ais (ite) substitute —C for F in (ii.) ; 5 “, C==(—C—32 af 32) OK ee) ene 32) = —5 C—160; 9 C+5 C= -—160 14 C=—160 c=- = ing, Conversion of Barometer Scales. — In this country the barometric pressure is frequently stated in inches instead of millimetres. The ordinary atmospheric pressure is 760 mm.=29'92 inches; and 26 AIDS TO THE MATHEMATICS OF HYGIENE one may be converted into the other by a simple rule of three—e.g., find the value of 735 mm. in inches: 760 mm, : 735 mm.: : 29°92 inches : x inches. E29 92k Dele a es 289 inches, The Centimetre-Gramme-Second System. Since May 1, 1914, the metric system has been adopted for meteorological measurements. In this system the cendzmetre is the unit of length, the gramme the unit of mass, and the second the unit of time. This system of units is known as the centimetre- gramme-second (or C.G.S.) system. The unit of velocity=1 centimetre per second. The unit of force (or the force which produces an acceleration of I unit. of velocity per second in a mass of I gramme) is known as a dyne. The untt of pressure=a dyne per square centimetre. 1,000,000 dynes per square centimetre=a megadyne. As a dyne per square centimetre is an exceedingly small unit, the megadyne per square centimetre has been adopted as the practical unit of atmospheric pressure in the C.G.S. system, and is known as a dar ; and barome- tric readings are now given in terms of cendtibars (745 of a bar) and mzl/ébars (7/55 Of a bar). The megadyne per square centimetre is equivalent to a pressure of 750°I mm., or 29°53 inches of mercury ; ss 29°53 inches=1I megadyne per square centimetre= 1 bar =1,000 millibars. Normal atmospheric pressure (760 mm., or 29°92 inches) 29°92 X 1,000 7 would thus be : 29°53 = 1,013°2 millibars, HYGROMETRY | 2% Relationship between the Different Varieties of Barometers. GLYCERINE BAROMETER. Example: If the height of a mercurial barometer be 30 inches, calculate what would be the height of a barometer made with glycerine. (Specific gravity of mercury=13°6, and glycerine = 1°28.)* Since the height of the column varies inversely as the density, . fheight of) , fheight of) .. fsp. gr. of {spa ae *"(mercury f ° (glycerine f °° (glycerinef ° (mercury J’ On aa : 1°28 EE 3:6:5 EC OCG aimee Pres afi Og BO inches= 26'6 feet. WATER BAROMETER. Example : From what depth is it theoretically possible—with the barometer standing at 30 inches—to raise water with a simple lift-pump ?* 3 The water rises in the pipe till the pressure of the liquid column counterbalances the atmospheric pressure on the water of the reservoir. The pipe is thus, prac- tically, a water barometer. The question, therefore, resolves itself into the following : What is the height of a water barometer, when a mercurial one stands at 30 inches ? a Since height varies inversely as density, therefore— * D.P.H. Exam, (Roy, Coll, Phys. Surg.). 28 AIDS TO THE MATHEMATICS OF HYGIENE height of) . fheightof) .. f{ sp.gr.) . { sp. gr.of), jee ot ; | water } ate joey p | mercury f- 20 seeds say et x= 13°6 X 30 = 408 inches = 34 feet. (In practice, water cannot be raised with a lift-pump more than about 28 feet, owing to the impossibility of obtaining a perfect vacuum.) Example : The lower valve of a pump is 30 feet 4 inches above the surface of the water to be raised; find the height of the barometer when the pump ceases to work. [Ans.: 26°76 inches. ] ALCOHOL BAROMETER. Example: A mercurial barometer falls 1 inch; what is the cor- responding fall in an alcohol barometer ? If the mercury stands at 30 inches, then the height of the alcohol barometer can be found as follows : fenctent te enue of). : {SP gr. of |, Jsp.gt oBl 4 mercury alcohol J** \ alcohol J * mercury J ’ SUC TeR Ye DSTORE aE Reh 20k 156 _ 510 inches, That is, 30 inches mercury = 510 inches alcohol ; : 10 : *. I inch mercury= = =17 inches alcohol. The alcohol barometer will therefore fall 17 inches. Correction of Barometric Height for Altitude. As a rule, the barometer falls 1 inch in ascending goo feet. If a=number of feet above sea-level of station HYGROMETRY 23 where observation is taken, then, in ascending that height, the barometer will fall ,45=o’oo! inch for every foot, or o’oolx inch in ~ feet. In correcting for altitude, therefore, o‘oo1x inch must be added to the reading taken. That is, Barometric reading ae senclevel =observed height +o°oo1z inch. For strict accuracy, the difference in the temperature of the air at the elevated station, and at sea-level, should be taken into account, but for ordinary observations the temperature is assumed to be that of the external air at the station where the barometer is placed. To Ascertain the Altitude by the Barometer. If x=barometric height (in inches) at lower station, and y=barometric height (in inches) at upper station, then (x — y) inches = difference between the two readings— z.e. (x—y) inch represents the fall. , Since barometer falls 1 inch for every goo feet ascended, it will fall (#—y) inches in goo (#—,y) feet. That is, there is a difference of g00 (x~y) feet between the two stations. To find the altitude, therefore, take barometric readings at both stations, and multiply their difference (in inches) by goo. The product will be the altitude (in feet). Example : Barometer at lower station = - - 29°92 inches. Barometer at upper station= ~ me BOR2 Too. Dierence =O 701s .. Altitude of upper station =900 X 0°71 == 639 feet. 30 AIDS TO THE MATHEMATICS OF HYGIENE Correction of Barometric Height for Temperature. It is usual to reduce all barometric readings to the freezing-point, in order to render them comparable in different places and at different times. Let a=coefficient of absolute expansion of mercury, == O00 1 MOr GVely. tart Let A=height at 32° F., and h:=height at 2° F. If volume of mercury at 32° F.=1, then volume at 7° F.=1-+ a(¢— 32). And since height of column varies inversely as the density, and density inversely as the volume ; therefore the height must vary See! as the volume, and there- fore he I ; Jus 1+a(¢— 32) ° ry ne ee ee a a2), This equation may be further simplified as follows : Multiply both numerator and denominator by [1 -a(¢—32)]; thus: h{t —a(¢—32)] ~ [1 +a(¢= 32)] [1 — a(¢— 32)] _Adi -a(¢- 32)] 1 —[a(¢— 32)?” Now a? is such an infinitely small number that the ex- pression [a(¢—32)]? may be neglected without in any appreciable degree affecting the result; the equation therefore becomes : h=h, [1 —a(¢—32)]5 DUti¢=0'O001, *, A=/y [1 —o’ooo! (¢—32)]. HYGROMETRY 31 [This correction is simply for the expansion of the mercury due to heat, and does not correct for the ex- pansion of the brass-scale.] Example: A mercurial barometer reads 30'1 inches at 200 feet above sea-level, and the attached thermometer 65° F. Make the necessary corrections for the correct reading of the barometer at sea-level and 32° F. (Coefficient of ex- pansion of mercury o’ooo! per degree Fahrenheit.)* Since the barometer falls 1 inch in ascending goo feet, ee 200 : ; it will fall wore inch in ascending 200 feet ; 0'22 inch must, therefore, be added to the observed height in order to correct for altitude. Thus -30°1 + 0°22 = 30°32 inches. Again, | h=h; [i —o*ooor (¢~ 32)] = 30°32 [1 —o'0001 (65 — 32)] = 30°32 [1 — (00001 x 33)] = 30°32 [1 -0'0033] = 30°32 X 0'9967 i= 30:22, Corrected reading = 30°22° F, THE VERNIER. Example : State the principle on which the vernier is con- structed.t The principle of the vernier is as follows: * -D.P:H, Exar. (Roy. Coll, Phys, Surg.). ¢ D.P.H. Exam., Cambridge, 32 AIDS TO THE MATHEMATICS OF HYGIENE A given length of the brass scale is taken, containing nm divisions; the same length of the vernier is taken and divided into +1 divisions—that is, the given length contains 7 divisions on the fixed scale, and #+1 divisions on the vernier. If S=length of one division on the scale, and V=length of one division on the vernier, then (z+1)V=2S ; © - teas. If each scale division =~ inch, then S=2; n I n and V=—— x —= —-—_,, inch, PE coy. ~ 3(a+1) Ij n and S-V= inch, x x(a+!1) Thee eae +1) We thus obtain the following : ts 1 Length of scale-division =~ inch, on ast Vernier | 4; inch, and - seyret bie I) difference between S and V division= ts oar inch. A. barometer scale is usually divided into inches, tenths of inches, and twentieths of inches—that is, the { eee nah Daa distance between the smallest divisions is 5 inch, or a In graduating the vernier, it is usual to take, for the length, 24 scale-divisions, ze, 2=24; then an equal length of the vernier is divided into 7+1=25 divisions. A vernier division will thus be less than a scale- division by =0°'002 inch, I Lee ae HYGROMETRY 33 Therefore, in reading a barometer, the number of divisions read off on the vernier must be multiplied by 0002, and the product added to the height already observed on the brass scale. The usual construction of the vernier has been here taken ; sometimes, however, z divisions on the scale are divided into #—1 divisions on the vernier; in other words, a certain length containing z# divisions on the scale may be divided into 7-1 divisions on the vernicr. (Sy) CALA be Tele ak HYDROSTATICS AND HEAT SPECIFIC HEAT. DEFINE the term ‘specific heat.’ How is the specific heat of bodies determined ?* The specific heat (S.H.) of a body=the quantity of heat which it absorbs when its temperature rises through a given range of temperature, compared with the quantity of heat which would be absorbed under the same circum- stances by the same weight of water, the specific heat of water being taken as I. A thermal unit (T.U.)=the amount of heat required to raise one unit of weight (¢.g., 1 lb.) of water through one unit of temperature (¢.g., 1° C.). 1f.S:H.=1; then 1 1b. raised 1°'C, =... Us [iS ie sat5e ss ae lbs as, Sion le TEStHe==27, so PDS; 4 oss Ore ee LiioelL «mesh csp ee DLDSa 0) tees enone Or see That is, the quantity of heat absorbed when a body is heated through a certain range of temperature = weight x temperature x specific heat. Similarly, when the body * D.P,.H. Exam, (Roy, Coll. Phys, Surg.), 34 HYDROSTATICS AND HEAT 35 cools down again to its original temperature, it parts with the same number of thermal units. The specific heat of bodies may be determined by (i.) the method of mixtures, or (ii.) the method of melting ice, both of which are sufficiently explained by the two following examples ; (i.) Method of Mixtures. Example: A piece of iron, weighing 20 ounces at a temperature of 98° C., is immersed in 60 ounces of water at a tempera- ture of 17° C. After the temperatures have become uniform, that of the cooling-water is found to be 20° C. What is the specific heat of the iron ?* Let c=specific heat of iron. The heated iron, weighing 20 ounces, has cooled from 98° C. to 20° C.—that is, through 78 degrees—and has, therefore, lost 20x78 xc thermal units. The water, whose weight is 60 ounces, and whose specific heat is 1, has been raised from 17° C. to 20° C.—d.e., through 3 degrees—and has, therefore, absorbed 60 x 3 x 1 thermal units. The number of thermal units lost by the iron must be the same as the number absorbed by the water. That is (heat lost)=(heat gained) ; “S80 78 XC = OOK 3% TS *, 1560¢= 180 eats 180 LB BE ea es Therefore specific heat of iron=o'1153, which means that the same quantity of heat which would raise 1 pound * D.P,H. Exam, (Roy. Coll. Phys. Surg.), 36 AIDS TO THE MATHEMATICS OF HYGIENE of iron through 1° C., would only raise 0°1153 pound of water through 1° C. (ii.) Method of Melting Ice. Example : A hole is made in a block of ice at 0° C., and an iron ball at 100° C., and weighing 200 grammes, is dropped in. When the ice ceases to melt the water formed is weighed, and found to be 29'2 grammes. What is the specific heat of the iron? Let c=specific heat of iron. The latent heat of water is 79 (vzde below)—that is, a gramme of ice at o° C. in melting to water at o° C. absorbs 79 units; and, therefore, 29°2 grammes will absorb 29'2 X 79 units. The heat given out by 200 grammes of iron in cooling from 100° C. to 0° C. will be 200 x 100 X¢ units. And since— (heat lost)=(heat gained), e200 X 100 X C= 29'2.%70 20,0006 = 2, 307 2,30 = 7307 ot 153. 20,000 LATENT HEAT. The latent heat of a substance is the heat absorbed (or given out) by a unit of a body in changing its state or condition (e.g., from ice at o° C. to water at o° C.). Example : 10 grammes of ice are mixed with 50 grammes of water at 29° C.; the resulting temperature is 11° C. Find the latent heat of water. Let L=latent heat of water. Then, heat required to melt Io grammes of ice into HYDROSTATICS AND HEAT . 37 water at o° C.=10 x L, and heat required to raise melted ice from o° C. to 11° C.=10X II=TII0 units 5 *, total heat absorbed by ice=(10L+ 110) units. Heat lost by 50 grammes of water in falling from 29° C. to 11° C. (@-e., through 18 degrees) =50 X 18=900 units. (heat gained) = (heat lost). e's IOL+ I110=900 10L=900— 110=7g0 E=79 That is, the change of 1 gramme of ice into I gramme of water at the same temperature, requires as much heat as will raise 1 gramme of water through 79° C. SPECIFIC GRAVITY. The specific gravity of a body is the number which expresses the relationship between the weight of a given volume of this body and the weight of the same volume of distilled water at 4° C. That is: _ weight of a fixed vol. (in air) ~ weight of equal vol. of water’ A. Solids. The ‘principle of Archimedes’ affirms that a body immersed in a liquid loses a part of its weight equal to the weight of the displaced liquid. But the volume of this displaced liquid is, obviously, the volume of the body which displaces it; therefore it follows that a body im- mersed in a liquid loses weight equal to the weight of its own volume of the liquid. If this loss of weight be ascer- _tained, then the weight of an equal volume of the liquid 38 AIDS TO THE MATHEMATICS OF HYGIENE is also known. Therefore, the specific gravity of a body may be represented thus: es weight of the body in air ~ loss of weight when weighed in water’ [For strict accuracy, the weighing should be performed 7m vacuo, to obviate the necessity of making any correc- tion for the weights of the unequal volumes of air displaced by the substance in one scale-pan, and the weights in the other. | B. Liquids. A body is weighed successively in air, water, and the liquid whose specific gravity is to be ascertained. The loss of weight when weighed in water=weight of an equal volume of water. The loss of weight when weighed in the liquid = weight of an equal volume of the liquid ; and since by the definition— weight of a vol. of the liquid. weight of an equal vol. of water ’ S.G. of liquid= loss of weight when weighed in the liquid “+ S-G.= Toss of weight when weighed in water Example : Six cubic inches of zinc weigh 24°8 ounces. What is the specific gravity of zinc ? 1,728 cub. in. (or 1 cub. ft.) of water weigh 1,009 oz. ; : 1,000 x 6 1,728 and 6 cub. in. of zinc weigh 24°8 oz. ; , o.G. Brie =—7SE8. 3°472 *. 6 cub. in. weigh — Sais AT 2 OZ, HYDROSTATICS AND HEAT 39 Example: What is the specific gravity of a body of which 2 cubic feet weigh a lbs? < 1,000 16 Ibs., 1 cub. ft. of water weighs 1,000 oz.= Ibs. ; 1,000 7 16 and # cub. ft. of the body weigh x lbs., “. 2 cub. ft. of water weigh x XX16 o'O16 x ae S.G.= — <= = ' 1,000 2 1,000 # n 16 Example: An ivory ball weighs 8°53 grammes in air, 4°07 grammes in water, and 3°93 grammes in milk. Find the specific gravity of (i.) the ivory, (i1.) the milk. (i.) In water the ball loses 8°53 — 4:07 = 4°46 grammes ; 8°53 .. S.G. of ivory = —==1'9I. Ivory 4°46 9 (ii.) In water the ball loses 4°46 grammes ; .. 4°46 grammes=weight of an equal vol. of water. In milk the ball loses 8°53 —- 3°93 =4'6 grammes ; .. 46 grammes= weight of an equal vol. of milk. 5 4°6 .. 5.G, of milk=— =1'032 4°46 : (or, in the more usual notation, 1032). Example : A piece of copper weighs ro pounds in air and 84 pounds in water. Find its specific gravity and its volume in cubic inches (a cubic foot of water weighing 1,000 ounces). 40 AIDS TO THE MATHEMATICS OF HYGIENE 10— 8f=14 lbs.=18 oz.=loss of weight in water. 1,000 oz.=1 cub. ft.=1,728 cub. in. ; el 20a *, 18 oz. = —————_ 5311 cub. in. 1,000 Since loss of weight=18 oz., . weight of displaced water= 18 oz. ; .. vol. of displaced water =31°I cub. in.; “; vol. Of piece of copper (f= 3171 cub.an.; weight in air loss of weight when weighed in water and S.G. of copper = One c.c. of water weighs I gramme; the specific gravity of copper is 8°9—that is, a piece of copper is 8°9 times heavier than the same volume of water; therefore I c.c. of copper weighs 8:9 grammes. Similarly, the specific gravity of lead is 11°347; therefore 1 c.c. of lead weighs 11°347 grammes. For convenience of calcu- lation, therefore, it may be remembered that ¢he weight of 1 ¢.c. of any substance ts tts specific gravity expressed im grammes. Example: A piece of wood, the specific gravity of which is 0°6, weighs 7” vacuo Ic grammes. To it is attached a sinker weighing 27 vacuo 13°3 grammes ; the specific gravity of the sinker is 11°347. What volume of water, expressed HYDROSTATICS AND HEAT 41 in cubic centimetres, will the two bodies displace when immersed in that fluid ?* Firstly, with regard to the wood: o°6 grm. wood=I c.c. O I ss 10 grms. wood = —-; eae 16°7 ¢.e. Since the volume of the wood measures 16°7 c.c., it will, when immersed, displace 16°7 c.c. of water. Similarly, for the sinker : 11°347 grms. sinker=I c.c, é 12 ‘o) 123g hlns; sinker = at] CG That is, the sinker, when immersed, will displace £17 c.c. of water: Total water displaced by wood and sinker will there- fore be; 167 Ver 7 17°87 Cc. As a corollary to the above problem, the size of the piece of wood required to support the sinker in the water may be calculated. When a body is immersed in a fluid, the upward pressure on it is equal to the weight of fluid displaced by the body. The downward pressure, due to gravity, is equal to the weight of the body. Where there is equilibrium, these opposing forces must be equal to each other. * D.P.H. Exam (Roy, Coll. Phys. Surg.), 42 AIDS TO THE MATHEMATICS OF HYGIENE Weight of sinker= 13°3 grammes, .. downward pressure on sinker = 13°3 grammes. Volume of sinker = 1°17 ¢.c., 7. vol. of displaced water— 1117.<.c., .. weight of displaced water = 1°17 grammes, .“. upward pressure on sinker = 1°17 grammes. Let x = number of c.c. of wood required. Then x c.c. wood weigh 0°6 x grammes; that is, weight of wood = 06 x grammes ; «. downward pressure on wood = 0°6 + grammes. Vol. of woodi==24 Cc .. vol. of displaced water = 4% c.c. ; .. weight of displaced water = x grammes; .. upward pressure on wood = x grammes. And since (upward pressure) = (downward pressure) ; ee LI7+4% = 1334004 x—0'6 = 13°3-T'17 OA wel 2s 2 ROO Therefore, wood required to float the sinker = 30°3 c.c., or 30°3 Xo'6 = 18°18 grammes. HYDRAULIC PRESS. Pascal’s law states that ‘pressure exerted anywhere upon a mass of liquid is transmitted undiminished in all directions, and acts with the same force on all equal surfaces, and in a direction at right angles to those surfaces.’ Example : The smailer piston of a Bramah’s Hydraulic Press is 4 inch in diameter, the larger piston is Io inches in diameter; a weight of 12 Ibs. is placed on the small. What will be the weight placed on the larger ?* * D.P.H. Exam. (Roy. Coll. Phys. Surg.), HYDROSTATICS AND HEAT 43 Area of circle=7 x ad (p. 178). Diameter of small piston = inch ; .*. area of small piston =/ x (3). Diameter of large piston= Io inches 3 *, area of large cu x'CEO)? area of area of re " "small piston large piston (3) a (10) a ae : (4)? : (10)? eae : 100 aa | > 400 ; That is, the area of the large piston is 400 times that of the small. The surface of the large piston may thus be considered to be made up of 400 units of surface, each unit having the same area as the small piston. And since a weight of 12 lbs. is placed upon the small piston, this weight will be transmitted undiminished (according to Pascal's law) to the water, and to each of the 4oo units on the surface of the large one. The pressure on the larger piston, therefore, will be— 12 X 400= 4,800 lbs.=2'14 tons. CHAPTER IV VENTILATION THE greater the velocity of the air, the greater will be the outflow, or amount delivered, in a given time; that is, velocity varies directly as outflow. Also, velocity varies inversely as the sectional area. This is not always obvious at first, but on a little con- sideration it will be seen to be true. Take, by way of example, a narrow river suddenly widening its bed for a distance, and then narrowing again. Exactly the same quantity of water which enters the widened portion must leave it (for if more entered than left, the water would be dammed back ; if more left than entered, the widened portion would run dry). Therefore, the same amount of water must pass the widened portion as passes an equal area of the narrow portion in the same time; but as the bed is widened, the velocity must be diminished in order to fulfil this condition. So the greater the sectional area, the less need the velocity be, in order to deliver a given quantity in a given time; on the other hand, the less the sectional area, the greater must be the velocity, in order to deliver the required amount in the time. Therefore, since the velocity varies directly as the outflow, and inversely as the sectional area, it may be represented as follows : 44 VENTILATION | 45 Velocity = __ outflow sectional area’ or, more briefly, he WO In the Delivery of Air through an Inlet, having given Two of the Three Following Data (viz., Velocity, Delivery, and Sectional Area), to find the Third. Examples : (i.) Find the velocity required to deliver 3,000 cubic feet of air per hour through an inlet, whose sectional area is 24 square inches. O = 3,000 cubic feet per hour, and S=24 square inches, ope tk SF square foct. 144 6 = = = aoa eae 18,000 feet per hour, 18,000 60 x 60 (or 3°4 miles per hour). =5 feet per second (ii.) Find sectional area of inlet required to deliver 3,000 cubic feet of air per hour, with a velocity of 5 feet per second. O = 3,000 cubic feet per hour. V=5 feet per second, or 18,000 feet per hour. O ae O _ 3,000 cubic feet _ I ee V___18,000 feet — 144 6 Oe 6 Square foot, = 24 square inches, 46 AIDS TO THE MATHEMATICS OF HYGIENE (iii.) An inlet, having a sectional area of 24 square inches, delivers air at the rate of 5 feet per second ; find total delivery per hour. V=5 feet per second. S=24 square inches= oe square foot. V=5 oh. ee Vaxts = 5 aleet xe square foot; = 2 cubic foot per second; _5x60x 60 __ {3,000 cubic feet ay 6 is per hour. For the sake of simplicity, the same figures have been taken in each of the above three examples, from which it is seen that an inlet whose area is 24 square inches will deliver 3,000 cubic feet of air per hour, with a velocity of 5 feet per second. There are one or two points to notice in the solution of the above examples. In (i.) and (ili.) one of the data given is in z#ches, the other in feet. These must, of course, be both brought either to inches or feet before proceeding further. In (ii.) O=feet per Zour, V=feet per second. Here, one of them must be brought to the same terms as the other. Also it may be pointed out that : cubic feet _¢ _ cubic feet square feet > feet square feet feet = square feet ; =feet ; feet x square feet=cubic feet. It will be noticed that no allowance has been made for friction (for further reference to this subject, vzde p. 61). It is usual, for an ordinary inlet, to deduct one-fourth for friction, VENTILATION 47 Thus, in example (i.) 5 feet per second becomes 5 —$=3'75 feet per second; and this lessening of the velocity will lessen the total delivery by one-fourth, which would then be only 3,000 - ao 242 50 cubic feet - per hour. With the lessened velocity, the same-sized inlet (viz., 24 square inches) would suffice to give the proportionately diminished delivery ; but, in order to produce the 3,000 cubic feet per hour, allowing the velocity to be diminished by friction, the inlet would have to be enlarged in the same proportion as the increased delivery required ; thus, 2250: 3000 3? 24°Sqrin. 24 Sq, In, : from which we find that x= 32 square inches. The above methods do not take into consideration the difference (if any) in the temperature of the inside and outside air, nor yet the difference in level between inlet and outlet. (In this case the velocity must be calculated by Montgolfier’s formula, p. 56.) The following is a simple method of calculating the total delivery and velocity of the air, and the sectional . area of the inlet. If the sectional area be 1 square foot and the velocity I foot per second, the delivery must obviously be 1 cubic foot per second. Sectional Area of Opening. Veloctty. Delivery. 1$q. it., ) I cub. ft. per sec., or 144 sq. in. per hour. 5 X 3,600 = 18,000 cub. ft. per hour. 18,000 _ {3,000 cub. Oo” th per Dr, V ft. per’sec.. ~-. or 3,600 cub. ft. ne | 1 ditto: ... .SefisperseG «> Ei =2d sq. in, ditto oes 48 AIDS TO THE MATHEMATICS OF HYGIENE To Calculate the Amount of CO, expired by an Adult per Hour. Inspired air contains 4 parts per 10,000, or 0°04 per cent. of CO, Expired air contains 4'o4 per cent. ; .. 4'04—0°04=4 per cent. of CO, is given off in each breath. In the case of an adult male, at each breath 30°5 cubic inches of air pass in and out of the lungs, containing, when expired, an additional 4 per cent., X 30° Ae. ; 4% 30°) — 1-22 cubic inches CO, 17 respirations per minute, or 1,020 respirations per hour, would produce 1°22 X 1020 =1244°4 cubic inches per hour, 1244°4 1728 =0'72 cubic foot per hour. ? Women and children exhale less than this, and 06 cubic foot per hour is about the average per head for a mixed assembly of people. To Calculate the Necessary Air-supply and the Impurity Present. Let D=delivery of air (in cubic feet), or amount of air available. E=total amount of CO, exhaled. r=added respiratory impurity in I cubic foot of air. VENTILATION 49 Then, added im- { “of | total | total de- purity in ¢ : 1 ft. of purity livery of I cub. ft. J air. | added ( air. Sueek eee Ds p= a r The following data should be borne in mind: On an average, each person exhales o°6 cubic foot of CO, per hour, and this figure is usually taken in all calculations. The total CO, in a room should not exceed 0'6 part per I,000; and since atmospheric air contains 0°4 part per 1,000, it follows that the amount of added respiratory impurity should not exceed 0°6 —0°4=0'2 part per I,0co. How much fresh air should each person be allowed per hour, in order that the above conditions may be fulfilled ? E =0'6 cubic foot per hour ; v=0'2 per 1,000, or 0'0002 per cubic foot ; i o'6 = —- = ——— = 3,000. r 0'0002 That is, each person should be allowed 3,000 cubic feet of fresh air per hour. Example : In a room 20 feet longx15 feet widex8 feet high, occupied by three persons, state how many cubic feet of fresh air per hour would be required; and what should be the aggregate dimens ons of inlets, so that the rate of in-flow of air should not exceed four linear feet per second.* * D.P.H. Exam., Cambridge, 50 AIDS TO THE MATHEMATICS OF HYGIENE Room = 20 X 15 X 8=2,400 cub. feet. Three persons exhale 0°6 xX 3=1°8 cub. feet CO, per hour. Permissible added impurity =0'0002 cub. feet. 19, 1'°8 18,000 ¥ 0100022 (For the first hour 9,000 — 2,400= 6,600 cub. feet.) 4 feet per sec.=240 feet per min, = 14,400 feet per hour. =G= ae = 2-sq. fect — a= go sq. inches. So, 9,000 cub. feet of air will be required per hour, and go sq. inches of inlet space. =9,000 cub. feet per hour. Example : The air of a room containing 20,000 cubic feet, in which 10 persons have been working for 5 hours, is found to contain 10 parts of carbonic acid in 10,000 parts. How much fresh air is entering per head per hour ?* Air of room yields 10 parts CO, per 10,000; CO, naturally present in air=4 parts per 10,000; therefore the added respiratory impurity = 10 — 4 =6 parts per 10,000, or 0'0006 cubic foot of CO, per cubic foot of air; that is, xr =0°0000. Ten persons exhale 10x 0'6=6 cubic feet per hour, or 30 cubic feet in five hours ; os N= 30. ape pam e ~ ¢ 0'0006 or 10,000 per hour for the 1o persons— that is, 1,000 cubic feet per head per hour. = 50,000 for the five hours, * Di PH. Bbxam (hoy .Cous Proves noire), VENTILATION 51 Since total delivery per hour=10,000 cubic feet, and the room contains 20,000 cubic feet of space, there is sufficient air in the room for the first two hours, but afterwards 10,000 cubic feet per hour (or 1,000 cubic feet per héad) must be supplied in order to keep the CO, down to the stated limit of 10 parts per 10,000. Example : (i.) Six persons occupy a room of 5,000 cubic feet space continually for six hours; calculate the percentage of CO, present in the air at the end of the time, supposing 10,000 cubic feet of fresh air have been supplied per hour, (ii.) In what time would the permissible limit of im- purity be reached if there were no ventilation ? (i.) Six persons exhale 0°6 x 6=3'6 cubic feet CO, per hour, or 36 xX 6=21'6 cubic feet CO, in 6 hours ; ie eon lo 10,000 cubic feet of air supplied per hour = 60,000 cubic feet in 6 hours, which with the 5,000 cubic feet originally present=65,000 cubic feet ; Sa D =65,000. E EF) i216 . D==; i PF Br Giese Taree per cubic foot, or 0'033 per cent. of added impurity. Add to this the CO, originally present in air—viz., 0°04 per cent.— .. Total impurity =0'033+ 0'04=0'073 per cent. [Problems of this nature may also be solved without the use of the formula, as follows : 82 “AIDS 10 THE MATH EMATICSOF HYGIENE CO, originally present=o0'4 per 1,000, or 2 cubic feet per 5,000. CO, exhaled, as has been already seen=21°6 cubic feet. The fresh air added (10,000 per hour for 6 hours) = 60,000 cubic feet, which naturally contain 4 cubic feet CO, per 10,000, or 24 cubic feet per 60,000. Therefore total CO, at end of 6 hours= Originally present eee scone Added by respiration aut Be os Added in fresh air ie re 7 ik0) Total tf: ace 5 fergie &) The amount of air available has been seen to be 65,000 cubic feet. Therefore, totalamount of CO, present at end of 6 hours=47°6 cubic feet in 65,000 cubic feet of air, ES Nes = =—O-,O7 en peCrmacene 5) 65,000 73 p ] (ii.) In what time would the permissible limit be reached, if there were no ventilation ? The permissible total impurity is 0°6 per 1,000 (p. 49), or 0°6X 5 =3 cubic feet in the 5,000. The room already contains 0°4 per 1,000, or 0'4 X §= 2 cubic feet in the 5,000. Therefore, the permissible added impurity=3-2=1 cubic foot. How long will it take the six persons to add this 1 cubic foot ? In 1 hour 6 persons exhale 0°6 x 6=3'6 cubic feet CQ, ; in = hour they will add 1 cubic foot CO,, re) and — Nee =16°6 minutes, 3°6 3°6 VENTILATION : 53 [This result may also be obtained as follows : Let #=time (in hours) ; CO, in room =2 cubic feet ; Six men in x hours add 3°64 cubic feet $ *. Total CO, in x hours =(3°6%+ 2) cubic fect. Available air= 5,000 cubic feet, And permissible limit of impurity =6 per 10,000. Since this limit must not be surpassed, the total CO, and the available air must be to each other in the pro- portion of 6 to 10,000, or Total COW: available air + <6". 10,000; os POX 2. : 50007: 1.6 + 10,000; 3-2 _ 1 36 36 ocr = hours = 16°6 minutes. | ONO T ~ em Example : A room 5,000 cubic feet capacity is occupied by four people for 12 hours; the carbon dioxide was found to be 1°2 parts per 1,000. How much fresh air per hour has been supplied to each person ?* [Ans. : 750 cubic feet per head per hour.] Example: A room is ventilated by four equal inlets having an aggregate sectional area of 48 square inches, is occupied by two adult persons, and is lighted by one No. 5 bat’s- wing burner consuming 6°5 cubic feet of oxygen, and producing 2°8 cubic feet of CO, per hour. Find the velocity of the air-current per minute through each inlet needful to keep down the impurity to the permissible limit. .H. Exam. (Roy. Coll. Phys. Surg.). .H. Exam., Cambridge. 54 AIDS TO THE MATHEMATICS OF HYGIENE The necessary supply of air for the gas-burner should be regulated by the impurity it produces, rather than by the amount of oxygen it consumes. Two persons exhale 2x 0'6=1°2 cub. feet COg per hour. One gas-burner produces 2°'8 _,, 4 y Total CO,g=4’o cub. feet per hour. The permissible limit of added impurity =o°2 parts of COx, per 1,000 or 0’0002 per cubic foot. poe - 2 es Sgt eee 901600, T 70.0002 2 That is, 20,000 cub. feet of fresh air are required pr hour. Sectional area of the 4 tubes= 48 sq. in. =~ =4 sq. feet. v=o (p. 45) =—- = 60,000 linear feet per hour. 3 StI 000, WN Ue Woe per minute. [NoTE.—Although the velocity through each inlet is required, the estimated velocity—viz., 1,000 linear feet per minute—must not be divided by 4. The fact that there are four inlets simply means that (¢/ ¢he guestion of Jriction be neglected) one-fourth of the total air-supply will be delivered through each of the four inlets with the same velocity as the whole air-supply would be through : : ; O one inlet of four times the size. In one case V= S" and 1:0 O in the other V=*\—-=—= Hee pateed VENTILATION 55 IMPURITY PRODUCED BY ARTIFICIAL ILLUMINATION. Combustion of Oil. It is estimated that a lamp with one moderately good wick burns 154 grains of oil per hour. Oil (paraffin) contains about 85 per cent. of carbon. That is— 100 grs. oil contain 85 grs. C., and 154 grs. oil contain 85x 154 100 The equation C,+ 202=2COz, shows that 24 grs. C. pro- duce 88 grs. CO, ; 88 x 130°9 =130°9 grs. C, . 130°9 grs. C. produce - = 480 grs. COg; *, 154 grs. oil produce 480 grs. COs. Now, I cub. foot of dry air at 60° F. (the average tem- perature of a room) and 760 mm. weighs 66°85 X 4901 ics aya) 7 53027 grs. (p. 11). .. 1 cub. foot CO, (at same temperature and pressure) ; 36°27 X22) 6... weighs or = 815°3 grs. (p. 5). So, if 815°3 grs. COg=1 cub. foot, then 480 grs. COg= a =0°59 cub. feet. Therefore 154 grs. of oil pr Lae 0°59 cub. feet of COg. That is, the combustion of the oil in the lamp produces 0°59 cubic foot, or rather more than 4 cubic foot of CO, per hour. And since a human sie exhales 06 cubic foot per hour, it will be seen that an ordinary lamp pro- duces, each hour, about the same vitiation of the air, in respect to the CO,, as each occupant of the room. 56 AIDS TO THE MATHEMATICS OF HYGIENE Examyle : A room is fitted with four Tobin’s tubes, each having a sectional area of 50 square inches, and in the room are two persons and a paraffin lamp, wh ch burns 1 ounce of oil per hour. Find the velocity of the air-current necessary to keep the impurity down to 0o°6 part per 1,000. 154 grains oil produce 0°59 cubic foot COQxz (p. 55) : ‘. I ounce oil produces ae =1°6 cubic feet CO, per hour, Two persons exhale 2xo0G6=1'2 cubic feet CO, per hour ; .*. total CO, produced =1°6+ 1°2=2°8 cubic feet. Total permissible amount of CO,=6 per 10,000. Amount present in air=4 per 10,000. ‘*, permissible respiratory impurity = 6 — 4=2 per 10,000 ; or, 0°0002 per cubic foot. Hero Deseo ry OO0002 = 14,000 cubic feet per hour; AES = 3,500 cubic feet , each ‘ Tobin’ must deliver per hour. Now, V=c (p. 45) = oe cubic feet 3,500 cubic feet sosquareinches 50 iad square feet = 10,080 feet per hour, or 2°8 feet per second. [No deduction has been here made for friction. | MONTGOLFIER’S FORMULA. When bodies fall to the earth, we find that different bodies fall through equal spaces from rest, in a given time ; and the space fallen through varies as the square of the time. VENTILATION 57 Any deviation from this law is due to the resistance of the air. If V=velocity in feet per second, g=acceleration due to gravity, h=height fallen (in feet), then V7= 227%. The value of g increases from the equator to the poles ; in the latitude of London, g=32'19 feet per second. For practical purposes, however, we may consider the value of g to be constant. Suppose Z=5 miles = 26,400 feet, Then V?2= 2¢h = 2 X 32°19 x 26,400= 1,699,632, and V= ,/ 1,600,032 = 1,303 feet per second. If the atmosphere be assumed to be of uniform density throughout, it would extend upwards to a height of about five miles; and the velocity with which air enters a vacuum is the same as a body would acquire in falling through a corresponding height; therefore the velocity of air entering a vacuum=1I,303 feet per second. If the air be not entering a vacuum, but a space containing air ata different pressure, its velocity will depend on the differ- ence between their respective pressures. Usually these pressures cannot be directly observed, but must be inferred from the difference between their temperatures. The difference in pressure can be estimated by ascertaining the difference in level between inlet and outlet, and multi- plying this by the expansion of air due to the difference in temperature ; and this difference of pressure may be taken to represent the height fallen. The equation V?=2¢/ may be expressed as follows: 58 AIDS TO THE MATHEMATICS OF HYGIENE V4 =204==2:X 32°19 XA =64'38 XZ ¢ . V= 64°38 x J2=8'02x JA (approximately 8 ,/2). That is, the velocity in feet per second of falling bodies=8 x ,/height fallen. Example : : Assuming that two adults sleep in the room (whose dimensions are given in the example on p. 181), how much inlet and outlet area per head must be given to supply each person with 1,500 cubic feet of air hourly, supposing that the outside air temperature is 4o° F., and the internal air temperature is 60° F., and the height of the heated column of air is 20 feet ?* Since difference in level between inlet and outlet = 20 feet, and expansion of air due to increase of tempera- cha nie tee ture of (60—40=) 20 an -. height fallen=20x 22 =4°° feet, 491 491 400 _ 20 and V=8V/h= nae 1 = 7/22, feet persecond Mh eT APG a = 25,992 feet per hour. Deduct ¢ for friction, then 25,992 4 V=25,992 — = 19,494 feet per hour. Since delivery required=1,500 cubic feet per head per hour, and velocity=19,494 linear feet per hour, the sectional area of the inlet will be (p. 45) : ‘90. HP ebxan., Cambriage, VENTILATION 59 O _1,500 V neuog a BOO Say sq. inches per head, 19,494 and there ought also to be an outlet of equal size. The cubic capacity of the room—viz., 2,160 cubic feet -—need not be considered, since the necessary air-supply is given in the question, and therefore has not to be estimated either from the size of the room, or from the amount of impurity present. Example: Explain the formula— Velocity = /2¢ x height (T — 2) x o'002 with reference to an air extraction shaft. The temperature of the air in a perpendicular ventil- ation shaft of 2x14 feet section being 80° F., its length 30 feet, and the temperature of the external air 50° F., what quantity of air per hour would the shaft be capable _of delivering ?* This formula is in reality V=84/Z, which has been already explained, for +/2¢ has been shown to be 8; ‘height ’=difference in level between inlet and outlet 30 feet) ; T—Z=difference between internal and external temperature (30° F.) ; and o'002 aie approximately. “. Velocity =8+/ 30 x 30 x 0°002, Ha/a = 10°73 feet per second, ice 628 feet per hour. * 1D. Py Baim. a 60 AIDS TO THE MATHEMATICS OF HYGIENE Deduct ¢ for friction— . 38,628 — Slee = 28,971 feet per hour. Sectional area of shaft = 2 x 13=3 sq. feet ; O= V XS = 28,071. 300,91 3.Cub feet per hour. Example : (i.) A room 30 feet long, 20 feet wide, and 12 feet high, containing 3 gas-burners, is occupied by 5 persons. If there be no ventilation, in what time will the air have reached its permissible limit of impurity? (ii.) What delivery of air is required to prevent the permissible limit being exceeded? (iii.) If the only outlet be a chimney 15 feet high, and the temperature of the room 10° F. higher than the outside air, find the sectional area of the chimney required. (i.) Five persons exhale 5 x0'6=3 cubic feet CO, per hour. 1 cubic foot of gas produces about 0'6 cubic foot CO, ; a gas-burner consumes about 5 cubic feet of gas per hour; therefore each burner produces 5 x 0°6= 3 cubic feet CO, per hour, and 3 burners produce 9g cubic feet CO; *. total CO,=3+9=12 cubic feet per hour ; ee 12s Dike). The room = 30 X 20 X I12=7,200 cubic feet ; ep) = 7.200, pun ae = 0016 MDS 200 1: VENTILATION 61 If at the end of hour ~=0'00167, how long will it take 7 to reach the permissible limit of 00092? Let +s=number of minutes ; then 0°00167 : 0'0002 : : 60 minutes : + minutes. “2=7°2 minutes: fij\Dae= r O0002 (iii.) Velocity in feet per second=8 a/ BS “per second=15,g18 feet per hour. After pages + for friction, this is reduced to 11,940 feet per hour. = 60,000 cubic feet per hour. Seo (p. Vee oe squaie feet. 11,940 Therefore the chimney must have a sectional area of 5 square feet. FRICTION IN VENTILATION. Example: Compare the ventilation in the two rooms, A and B. The room A is ventilated by a straight shaft, circular in section, 30 feet long, with a sectional area of 1 square foot. The room B is ventilated by four shafts, square in section, each 30 feet long, and each having a sectional area of } square foot. It will be noticed that the sum of the sectional areas of the four smaller shafts in B is equal to the sectional area of the larger shaft in A, and the lengths are the same— viz., 30 feet. Therefore, the cubic contents, or capacity of the four shafts in B, are together equal to the capacity of the larger shaft in A, and would be capable, therefore, of supplying the same quantity of air to B as the single 62 AIDS TO THE MATHEMATICS OF HYGIENE shaft does to A, were it not for friction, which will now be considered. In two tubes of similar shape and equal sectional area, the friction will vary directly as the length—e.g., in two similar tubes, one 20 feet and the other 30 feet long, the friction in the longer tube will be increased by one-half, since its length is increased by one-half. If a shaft 40 feet long be increased + of its length (¢.e., become 50 feet long), it will have an aneteed friction of 1. In two tubes of different shape, but of ne same sectional area, the friction varies directly as the periphery. For example, take a circular and a square tube, each with a sectional area of I square foot. ‘The periphery of the circle, with a sectional area of 1 square foot or 144 square inches, may be found as follows (vzde p. 178) : 144 144 6 DR ee iW Seay oe Qahe meer, (pa We 7= 4/45°8305 =6'77 5 therefore, diameter =6'77 x 2=13'5 inches. ty2= FAAS. iN.5 os If diameter=13°5, then periphery =13°5 x 3°1416=42 inches, or 34 feet. A square with sectional area of 1 square foot has sides 1 foot long, and, therefore, periphery =4 feet. Thus ; Friction.in ) |) / friction: in) wees circular ae square tube f °° 32 4. To return to the comparison of the rooms : In A, shaft=30 feet; and periphery = 3% feet ; so total friction may be represented as : 30 X 3$=105 units. In B, the sectional area of the small square shafts VENTILATION 63 being 4 square foot, each side will be 4 foot long, and the total periphery of the four sides=4x4=2 feet. Each shaft = 30 feet in length, and periphery =2 feet ; .. total friction in each shaft = 30x 2=60 units, and total friction in the four shafts=60 x 4=240 units. So the following result is obtained : Friction in A: friction in B :: 105 : 240; ene Aaah 0) That is, the friction in B is more than twice that in A. If, in each shaft in B, we place a right-angled bend, the friction will be doubled ; or, Pe Bie 7. OX Ss 37:32; or friction in B is between four or five times that in A. Therefore, to get the same ventilation in B as in A, we must either increase the number of the smaller shafts, or the size of them, between four and five times if the bent tubes be used, or more than double them if the straight small shafts be used. Example : Suppose two ventilating shafts, the distance from the floor of the room to the point of exit being 50 feet in each case. Shaft A is straight, circular in section, with diameter of 8 inches. Shaft B is bent once at a right angle, cross-section is rectangular, measuring Io inches by 5 inches, and the outside temperature is 50° F.; the inside temperature in each case is 65° F. What are the respective volumes of air per hour which may be ex- pected to pass out of the shafts ?* ' D.P.H. Exam., Cambridge. 64 AIDS TO THE MATHEMATICS OF HYGIENE Since difference in level between inlet and outlet= 50 feet, and expansion of air for temperature BOS SS OMmLSs 49t 49v *, h=50x 75 — 1'5275, 491 and V=81/A=84/1'5275 =8 X 1'2359=9'8873 feet per second, or 35,494 feet per hour. Sect. area of A=a@?x 0°7854 (p. 178)=(8)7 xo'7854= 50 sq. inches = mi sq. feet. wo =V X 535.404 x Tae 12,324 cub. feet per hour. And sect. area of B= 10x 5=50 sq. inches. The sectional areas of A and B are thus equal in size, their lengths are the same, and both are exposed to the same internal and external temperatures; but owing to their different shapes, and the bend in B, the delivery of air through them will be differently affected by friction. The periphery (or circumference) of the circular shaft is 8 x 3°1416=25 inches. The periphery of the rectangular shaft having two sides of 10 inches, and two of 5 inches = 30 inches, .. Friction at _ f friction in ae BP Saco O: circular shaft angular shaft Bier UO: The right-angled bend in B doubles the friction in it. a ehaeictiondineA gs frictionein) Dems weno, Poe es Therefore, if A can deliver 12,324 cubic feet per hour, B will only deliver 27 tS 5,135 cubic feet per hour. eer, VENTILATION | 65 VENTILATION BY PROPULSION. To Calculate Delivery of Air by a Revolving Fan. The revolution of the fan sets the air in contact with it in motion. Each part of the fan is not, of course, revolv- ing with the same velocity, which varies from a maximum at the extremities to #z/ at the centre. The velocity of the air is usually taken to be 3 of that of the circumference of the fan, and is called the ‘effective velocity.’ So, if the speed of revolution of the extremity of the fan be known, the rate of movement of the air will be 3 of this. Example: Suppose fan=12 feet in diameter, then the circumfer- ence will be 12 x 3°1416= 37°70 feet (vide p.178). Therefore, in one complete revolution, the extremity of the fan travels 37°70 feet. If the fan be revolving 60 times per minute, the velocity of the extremity will be 37°70 x 60 = 2,262 feet per minute, and the ‘ effective ky 32,262 : , : velocity ’=~“———- = 1,696 feet per minute, or 101,760 feet per hour, which is the rate of movement of the air. If the sectional area of the conduit, through which the air is delivered into the room, be known, then the dis- charge in cubic feet can be at once calculated by the formula O=VxS (p. 45). By reversing this process, the size of the fan necessary to ventilate a room could be calculated, if the size of the conduit and the total delivery per hour of fresh air required were known. ‘These two data would give the velocity per hour of the entering air, since 3 ve ra ~) Ce a 66 AIDS TO THE MATHEMATICS OF HYGIENE Suppose this velocity to be 84,840 feet per hour ; then the rotatory velocity of the fan must be such as to impart to the air a rate of movement of 34,840 feet per hour. That is, the ‘ effective velocity ’=84,840 feet per hour, or ? of the velocity at the extremity of the fan. Therefore, Velocity at extremity of fan= 4 SbOHO 113,120 feet per hour, or 1,885 feet per minute. Now, if D=diameter of fan (in feet), then D x 3:1416=circumference of fan (in feet); and if R=number of revolutions per minute, then D x 31416 X R=distance travelled by circumference of fan per minute; but this=1,885 ; *. Dx 3°1416X R=1,885 ; SENT) SiR Meee eon ea norreete 31416 600 600 : D=5- Sand vk == D? 50, df = 5) thenws 1206 ifs = 3) thensk=-200; itp 2then hie 1300, That is to say,a fan of 5 feet diameter with 120 revolu- tions per minute—ov one of 3 feet diameter with 200 re- voluticns per minute ; 07 one of 2 feet diameter, having 300 revolutions per minute—would each of them give the required ventilation. CHAPTER V RAINFALL AND SEWERAGE To Calculate the Amount of Water-supply avail- able from Rainfall. (i.) Amount of roof-space. It is only necessary to ascertain the area of hori- zontal surface covered by the roof, the slope of the roof not affecting the result. Whatever its slope, the roof simply catches the rain which would have fallen on the area of horizontal ground, now covered by the roof, if the roof had not been there. (ii.) Average amount of rainfall per annum is about 30 inches. (iii.) The loss by evaporation is about one-fifth of the total rainfall. Example : The roof-space in a town is 60 square feet per head, and the annual rainfall is 30 inches. Find the amount of water available per head per annum. Roof-space = 60 square feet; rainfall = 30 inches= 24 feet ; 67 68 AIDS TO THE MATHEMATICS OF HYGIENE . amount of water per head per annum=60x 24= 150 cubic feet. Deduct = = 30 for evaporation ; .*. available water = 150 — 30=120 cubic feet, = 120 X 6'23=748 gallons. So the amount of water available for each individual during the year is 748 gallons, or about 2 gallons per diem. Example : How much water will 1 inch of rain deliver on 1 acre of ground ? One inch (or .; foot) of rain over I square yard (or 9 square feet) will give 9x + —3 cubic foot of water ; 1274 = eh One acre will give 4,840 x 4°6725 =22,615 gallons, = 22,615 X 1O=226,150 pounds ; scald = 101 tons (nearly) 2,240 (or, approximately, ‘ Ioo tons per inch per acre’). [1 acre=4,840 square yards. I cubic foot = 6°23 gallons. 1 gallon=Io pounds, 2,240 pounds=1 ton.] x 6'23=4'6725 gallons. RAINFALL AND SEWERAGE © 69 General Formula for Rainfall. Let R=annual rainfall (in inches), E =annual evaporation (in inches) ; Then (R-E) inches, or RS fall. Let A=number of acres of collecting-ground, Or (A x 4,840 x 9) =number of square feet of ground. a feet=available rain- Total water=(A x 4,840 x 9) x R zs E cubic feet | = 3,630 Ax (R-—E) cubic feet, = 3,630 A x (R - E) x 6'23 gallons, = 22,615 x Ax(R-E) gallons per annum, _ 22,615 xAX(R-E) — 365 =62 A (R—E) gallons per diem. p) Hawksley’s Formula for Storage. Let D =number of days’ supply to be stored ; F =mean annual rainfall in inches ; Then D= seid JF In this formula, F may be taken to be either @ of the Wrean annual yield of several years, or the eer of the three driest consecutive years Example : A cottage occupied by five persons has a horizontal roof-space of 360 square feet, and the annual rainfall is ‘25 inches; find the amount in gallons of the available ‘supply per head per annum, and state in cubic feet the -size of a tank needful to satisfy Hawksley’s s formula.* = D. P, H. Exam., Cambridge. a a ee 70 AIDS TO THE MATHEMATICS OF HYGIENE Roof-space = 360 sq. feet; rainfall=25 inches = = feet. S Se *, amount of water per annum = 360 x = =750 Cub. 1eet. Deduct ea 150 for evaporation, *, available water =750—150=600 cub. feet. = 600 x 6'23 = 3,738 gallons, or HIS 748 gallons per head per annum. 1,000 1,000 “/ 25 5 must be stored. The available daily supply for the five persons is 600 — cub. feet, 365 *, 200 days’ supply = By Hawksley’s formula = 200 days’ supply 600 X 200 . e . 365 which is the size, therefore, of the necessary tank. = S2oiclb ice: FLOW IN SEWERS. Hydraulic Mean Depth. Example : What is meant by ‘Hydraulic Mean Depth’? What is it in the case of a circular pipe running half-full? Show that it is the same when the pipe runs full.* The Hydraulic Mean Depth (or Radius) in a sewer is the sectional area of the fluid, divided by the length of that portion of the circumference of the pipe which is in contact with the fluid; or, more briefly: ———s ee * D.P.H. Exam. (Roy. Coll. Phys. Surg,). RAINFALL AND SEWERAGE ee: H.MD. _ sectional area of fluid wetted perimeter (i.) In a sewer running full, the sectional. area of the fluid will be the same as that of the pipe, and the wetted perimeter will obviously be identical with the circumference ; (ii.) In a sewer running half-full, the sectional area of the fluid will be exactly half that of the pipe, and the wetted perimeter will similarly be one- half the circumference ; wf ae ee 2 That is, in a circular pipe running full or half-full, the H.M.D. is one-half the radius, or one-fourth the diameter. Eytelwein’s Formula. This is founded upon the observation that the mean velocity per second in sewers is very nearly nine-tenths of a mean proportion between the H.M.D. and the fall in feet in 2 miles. If D =hydraulic mean depth, and F=the fall in feet per mile, then 2F=fall in feet in 2 miles. And if += mean proportion between these two quantities, then— DY ose ae £°-=2)DF, and x= +/2DF, gx \/2DF fe) and nine-tenths of this will be 72 AIDS TO THE MATHEMATICS OF HYGIENE : DF Therefore, mean velocity per second = 2% V2DF, and mean velocity per minute af x 60=54°/2DF. Eytelwein’s formula gives | V=55/2DF where V=velocity in feet per minute, D =hydraulic mean depth in feet, F = fall in feet per mile. Having given Two of the Following—(a) Velocity of Flow ; (4) Diameter of Sewer; (c) Gradient—to Find the Third.. (i.) A circular house-drain, running full, has a diameter of 4 inches; the gradient is 64 feet to the mile. What is the velocity of the discharge ? Making use of the above formula, diameter __. = —— =1 inch= 5 foot ; 4 P= 64. o's V=55V2 x yb X64=55V 10°6=55 X 3°27 = 180 feet per minute, = 3 feet per second. (11.) Calculate the gradient of a 12-inch sewer required to maintain a velocity of 3 linear feet per second.* Diameter= 12 inches=! foot ; *\ D =F foot. * D.P.H.' Exam, Cambridge, RAINFALL AND SEWERAGE 73 Velocity =3 feet per second = 180 feet per minute ; oe 55 /2DF = 180, 55°\/2.4.F = 180; «a © -(2)'- 1,296 ‘g Xp ee a 121 OPA ial 5,280 246° *, Gradient required=1 in 246, or a fall of 21°4 feet per mile. (ii.) A circular drain is laid with a faJl of 64 feet per mile. What should be its diameter in order that, when running full, it may discharge at the rate of 3 feet per second? Let d=diameter required (in feet). F=64, and V=3 feet per second=180 feet per minute. Hydraulic mean depth= a therefore D ina iS pga fs . 1296 oe 55 eee a and d= Me feet = vat a Buslecley inches. 121 X 32 121 x 22 74 AIDS TO THE MATHEMATICS OF: HYGIENE Knowing the velocity and the sectional area, the dis- charge from the sewer can be calculated. Example : How much sewage would be discharged in 3 hours by a sewer, circular in shape, with a diameter of 18 inches and a fall of 1 in 180, running half-full ?* Diameter= 18 inches=3 feet ; rape” x a foot, Zenit 6 ; 5280 4 Fall=1 foot in 180 feet =~. feet in 5,280 feet, =29°3 feet per mile; 2 Rie 20/3. V=55 V2DF=55N2x 3x 29'3=55 V22=55 X47 =258°5 feet per minute. O=VxS (p. 45). 2 S= (3) x0'7854 (p. 178)=1°77 square feet ; »*. O= 2585 x 1°77=4,575 cubic feet per minute, = 274,524 cubic feet per hour, = 823,572 cubic feet in 3 hours ; or, 823,572 X 6'23 = 5,130,853 gallons in 3 hours. And half this amount—viz., 2,565,427 gallons—will repre- sent the discharge in 3 hours, when running half-full. Example : Over an area of 1,000 acres a rainfall of 4 inch per hour occurs. What must be the diameter of a drain to carry this away, the gradient being 10 feet per mile? The diameter can be ascertained from the sectional * D.P.H. Exam, (Roy. Coll. Phys. Surg. ). - RAINFALL AND SEWERAGE 715 area, and the sectional area may be found if the velocity and outflow are known, since S=9 (p. 45)3 therefore: (i.) Find the outflow. . 1,000 acres = 1,000 x 4,840 square yards, = 1,000 X 4,840 X 9 = 43,560,000 square feet ; i; I and j i foot ; -- Amount of rain falling = 43,560,000 x -7- = 605,000 cubic feet per hour ; or, outflow= 10,083 cubic feet per minute. (ii.) Fi.d the velocity of flow. Let d=diameter of drain required (in feet) ; V=s5V2DF; pD=£(p. V4) F S105 a ee 2 s V=55 J ox2x 10o=55Nsd—=55 x 2'236Va =122'98 /d feet per minute. (iii.) Find sectional area of drain. iO” 10083 a S198 Vi y22:98Vd Va (iv.) Find the diameter. S=d? x.0'7854 (p. 178) ; square feet. Sr Og ces 22 a "™ ~0°7854 0°7854x J/g’ — 81°98 e 2 / = e e “2 aexX Na = O78 e4 104°39 5 that is, d@? =104°39; ce 2 log d= log 104°39 = 2°0186589 5 eae dare eee =0°807 4636 = log 6°4 3 o. 2=6'4 feet. | No deduction has been made for loss by evaporation. ] 76 AIDS TO THE MATHEMATICS OF HYGIENE Example : How many gallons of water per minute would be dis- charged from a drain-pipe having a diameter of 2 inches and a slope of ‘1 in 10, using for the calculation the formula— V =1401/ HS —114/HS, where V=the number of linear feet issuing per second, H =the hydraulic mean depth, S=sine of the slope. (Tables of logarithms are provided.)* Draw any horizontal line AB, such that the length AB=10 units. From the point B draw BC perpen- dicular to AB, and of a length equal to 1 unit; join AC. Then AC represents the slope of the drain (1 in 10), and the sine of the slope is the sine of the angle BAC. The length of the line AC is ~/(AB)?+(BC)2. (Euclid I. 47.) = 1/(10)? + (1)? =+/Io1 = 10°05. The sine of an angle is given by the trigonometrical perpendicular | ratio hypotenuse ” ; BC I **. one pe AGE iG owe . The diameter of the drain=2 inches=} foot ; there- fore the hydraulic mean depth=+x4=3; feet. I Sa TiS oe se Dk ee toori 168 24 10'O5 ae24i 2 *. V=1407/0'004146—114/0'004146. * D.P.H. Exam., Cambridge, RAINFALL AND SEWERAGE 79 Now log. 140° 01004146 = log. 140+ 4 log. o'004146 ;s = 2'1461280+: goa ; = 2°1461280+ = doulas: ; = 0°9549420 ; =log. 9°0145 ; .*. 1404/0'0041 46 =9'0145. Similarly, it may be shown that— 11 4/0°004146=1'76713 Therefore V=9'0145 — 1'°7671 =7'2474 linear feet per second = 434'8 linear feet per minute. Diameter of drain=} foot ; *, Sect. area= (3)? x0°7854 (p. 178), 0°7854 36" 0'7854 36 sq. feet. . Outflow = 434'8 x ——— cub. feet per minute (p. 45), 4348 x 0'7854 x 6'23 36 This multiplication and division should be performed by logarithms as follows : gallons per minute. Log. outflow— =log. 434°8+ log. 0°7854+log. 6'23—log. 36, = 2°6382895 + 1°8950899 + 0°7944880— 1'556302:, = 1°7715649, = log. 59°097 5 *, Outflow = 59 gallons per minute. 78 AIDS TO THE MATHEMATICS OF HYGIENE The Relationship between the Diameter, Velocity, and Gradient of Sewers. Example : How does the fall of a sewer affect the velocity of flow aT itera If V=present velocity, and v=the velocity to be found, then V=55./2DF, and v=55/2D'F'; andV:v:2 55V2DF: 55V/2D'F’; AoE ase we Dirt Speeder ike heed BD ee 1ST fe rise BRS IDM ho but since the diameter does not vary, D=:D’, ARN EIR Sok eS 1 that is— (i.) Where the diameter is constant, the fall varies as , the square of the velocity. ix Similarly, it may be shown that— (ii.) Where the velocity is constant, the fall varies inversely as the diameter, or LRN A aie Mag 2 and— (iii.) Where the fall is constant, the diameter varies as the square of the velocity, or WP SES SRLS Val Example: Show by the formula V=554/H2F that sewer-pipes of the following diameters and gradients have each the same velocity of outflow : * D.P.H. Exam, (Roy. Coll. Phys. Surg.). RAINFALL 4ND SEWERAGE 79 (a) Diameter 5 feet, fall 4 feet per mile. (6) Diameter 2 feet, fall 10 feet per mile. — (c) Diameter 1 foot, fall 20 feet per mile.* Changing H (hydraulic mean depth) in the above formula to D, and so using the same notation as has been already adopted in this chapter, the formula becomes V=554/2DF. If a sewer of 5 feet diameter has a fall of 4 feet per mile, what must the fall be in the case of a sewer 2 feet in diameter to ensure the same velocity ? | AED seh ae eran *hcAe RB SAN 2 ers *. 2F =5xX4=20; is) ew fe) That is, to obtain in (4) the same velocity as in (a), the gradient in (46) must be one of Io feet per mile, which proves the proposition in the case of (6). Similarly for (¢) : | aa Seay a aa UBNGED lnseteee Sey (Acie = 25 Xs _ which proves (c). Example : Estimate the relative volumes of sewage discharged from the three sewers whose dimensions and gradients are given in the previous example. Sect. area= a? x 0°7854 (p. 178) ; . Sect. area of (2)=(5)?x0'7854; 99 ” (3) =(2)? x 0°7854; ” ” (¢)=(1)? x 0'7854. * D.P.H. Exam., Cambridge, 80 AIDS TO THE MATHEMATICS OF HYGIENE Thus, the sectional areas vary as the squares of the diameters. The gradients are such as to produce the same velocity in each, and v=e : therefore, since V is constant, the volume of sewage dis- charged (O) must vary directly as the sectional area (S), and therefore directly as the square of the diameter of the sewer. % And since the squares of the diameters of (a), (0), and (c) are in the proportion of 25, 4, and 1 respectively, the volumes of sewage discharged will be in the same ratio; that is, (2) will discharge 25 times, and (4) 4 times the volume that (c) will in a given time, and with equal velocity. EXCRETA. Estimation of the Amount of Nitrogen and Ammonia in Excreta. (1.) Ax average adult man passes daily : 4 ounces of feeces and 50 ounces of urine. 1 fluid ounce water (sp. gr. 1000) weighs 1 ounce = 437'5 grains; : : 37°5 X 1020 *, I fluid ounce urine (sp. gr. 1020)= 437 ieee - =446'25 grains ; .*. 50 fluid ounces = 446'25 X 50= 22,312 grains. Urine contains 4'2 per cent. solids ; S solids 22812442 paar “. solids = a 937'I grains. Of the solids, 54 per cent. is urea ; 937°1 x 54 .. Urea = +*———— = 506 grains 100 509 8 2 RAINFALL AND SEWERAGE 31 Urea—CO(NHg)2—contains 28 parts N in 60 parts : Wiay 500% 23 ‘ ne Os leery Pea =236 grains. 14 grains N are contained in 17 grains NH; ; 230X117 .*. 236 grains N= = 287 grains NH3. 14 Fzeces contains 23°4 per cent. solids ; “. 4 ounces (or 1,750 grains) contain eae = 409°5 grains solids. Of the solids, about 16 per cent. is N ; N= 4005 X 16 a0 =65°5 grains ; and NH,=S5%17_ 795 grains. Therefore, the daily excreta of an adult man contain Total N=236+65 = 301 grains. Total NH3;=287+79= 366 grains. (il.) Jz a mixed community (including women and children) the daily quantity per head averages : 2°5 ounces feces, and 40 ounces urine ; and the amount of N contained therein can be estimated in a precisely similar manner and with the following results : 40 ounces urine contain 189 grains N, or 229 grains NH; ; 2°5 ounces faeces contain 41 grains N, or 50 grains NH. Therefore, the daily excreta per head of a mzxed com- munity contain : 6 82 AIDS TO THE MATHEMATICS OF HYGIENE Total N=189+ 41 -=230 grains. Total NH,=229+ 50=279 grains. [These quantities, dealing with solids only, are water- free. | Example: How much solid and liquid excreta are passed by an average adult man per diem, and how much water-free solids does the amount represent ? How much per annum would a mixed community of 50,000 persons pass, and how much ammonia would be contained in the total bulk 2 * An average adult man passes daily : 4 ounces faeces and 50 ounces urine. Feeces contain 23°4 per cent. solids ; eran 3 ; .. 4 ounces contain 4~*34 =0'936 ounces solids. Urine contains 4°2 per cent. solids; : . 50X4'2 : .. 50 ounces contain ere =2°I ounces solids ; .. total water-free solids =0°936 + 2°1 = 3'036 ounces. In a mixed community the quantities per head per diem average : 2°5 ounces fzeces, and 40 ounces urine. Therefore, in a year, the feces of 50,000 persons would amount to: 2°5 X 365 X 50,000 ounces ; 2°5 X 365 X 50000 or ; 16 X 2240 = 1,273 tons; and the urine would amount to: 40 X 305 X 50000 _ TerOG = 20,368 tons. * DP oiy hsxams Cam bridge, RAINFALL AND SEWERAGE 83 Total excreta per annum = 1,273 + 20,368 = 21,641 tons. [These quantities are not water-free.] The average daily excreta of a mixed community per head has been seen (p. 82) to contain 279 grains of NH,; therefore, in a year, the excreta of 50,000 persons would contain : 279 X 365 X 50,000 grains ; 279 X 365 X 50,000 are x16% 2,040 = 325 tons NH3. or, If, in the first part of this example, the solids, instead of being water-free, had contained 25 per cent. of moisture, then the remaining 75 per cent. would represent the solids; that is: 3°036 ounces=75 per cent. of the whole, 3036 X 100 oe = 4'048 ounces=total weight. and CHAPTER VI ENERGY, EXERCISE, AND DIET THE fact that any agent is capable of doing work is usually expressed by saying that it possesses energy, and the quantity of energy it possesses is measured by the amount of work it cando. The‘ unit of work’ is generally taken to be the quantity of work which is done in lifting 1 pound through a height of 1 foot, and this quantity of work is called 1 ‘ foot-pound.’ The product of the weight lifted (expressed in pounds), into the height through which it is lifted (expressed in feet), gives the amount of work done (in foot-pounds). Thus, a weight of 20 pounds lifted through a distance of 1 foot=20 foot-pounds; or, a weight of 1 pound lifted through a distance of 20 feet = 20 foot-pounds. In the same way, the ‘work done’ by a man during exercise can be ascertained. If W=his weight (in pounds) and D=the vertical height (in feet) to which he lifts his body, then the work done= W x D foot-pounds ; or if D=height in wzz/es, then 5,280D =height in feed, and work done W x 5280D =W x 5,280D foot-pounds, or 2240 foot-Zons. So far the lifting of the man’s weight vertically upwards has only been considered. What will the work amount to 84 ENERGY, EXERCISE, AND DIET 85 if, instead, he propel his weight horizontally along level ground? A great deal will depend upon the velocity with which he walks. It has been calculated that at an ordinary rate of three miles per hour, a man, walking along level ground, does work equivalent to raising his own weight, vertically, through = the distance travelled ; or, what is I onset the same thing, raises 55 of his weight through the whole distance travelled. It has been seen that raising his whole weight through the distance D miles requires an expenditure £ W x 5280D 2240 weight through the distance D will only require an ex- idle Beco x + foot-tons ; which number, 2240 20 therefore, gives the work done in walking a distance of D miles on level ground, at the rate of three miles per hour. W has been taken to represent the total weight—z.c., weight of body+ weight carried. If W=weight of body, and W’=weight carried, then the formula becomes (W+W’) x 5280D 2240 §} 6 : ae . The fraction ae spoken of as the ‘ coefficient of : I ‘ foot-tons ; therefore, to raise ie of his penditure of I 14 x — foot-tons. 20 resistance’ (or traction), and varies with the rate of walking. At three miles per hour, on level ground, it is I , equivalent to (approximately =) at four miles= I 20°59 es —~, and at five miles= P S 16°75 14°10 e 86 AIDS TO THE MATHEMATICS OF HYGIENE If C=coefficient of resistance, then the formula may be written (W + W’) x 5280D 2240 Where W = man’s weight (in pounds). W’= weight carried ¥; D =distance walked (in miles). C=coefficient of resistance. x C foot-tons. EXERCISE. Example: A bricklayer’s labourer weighs 150 lbs. and carries a load of bricks—4o lbs.—up a perpendicular ladder 30 feet high 109 times a day. Calculate how much ex'ernal work he does daily, and what will it be equal to in miles walked on the flat at the rate of 3 miles an hour ?* a tes 150+ 40= 190 lbs.= Rear b) ton. This weight has to be lifted through a height of 30 feet 100 times a day, or through a total height of 3,000 feet ; 190 2,240 .. total work= X 3,000 = 254 foot-tons daily. Find the equivalent of this in miles walked on the flat at the rate of 3 miles an hour: (W+W)X5,280D 2,240 where W + W’=190, D =distance walked in miles, C=y53 total work=254 foot-tons ; CG: * D.P.H. Exam., Cambridge. ENERGY, EXERCISE, AND DIET 87 190 X 5,280D bf 2,240 a ee 254 X 20 X 2,240 and D= =11°3 miles. 190 X 5,280 For the perpendicular ladder substitute a plank 30 feet long, with a gradient of I in 12. Total distance walked = 3,000 feet. If the plank rises 1 foot in 12 feet it will rise b eaea == 250 feet 12 in the total distance walked. Therefore, in addition to walking 3,000 feet on level ground, he has to raise his whole weight through a height of 250 feet. The energy expended in this portion of work will be 190 ; —— X 250=21'2 foot-tons ; 2,240 and in traversing the distance on level ground at the rate of 3 miles an hour, the work done will be 190 2,240 X 3,000 X 5! = 12'7 foot-tons. “. Total work=21'2+ 12'7= 34 foot-tons. [NOTE.—The distance travelled on level ground is equivalent to raising 3/5 of the weight through the whole distance. Whilst the upward distance is equivalent to raising the whole weight through the vertical height ascended. An ‘ordinary day’s work’ is generally taken to be 300 foot-tons ; and this amount of.energy would be expended in walking 14°2 miles on level ground at the rate of 3 miles per hour.] 88 AIDS TO THE MATHEMATICS OF HYGIENE METHODS OF CALCULATING DIETS. Method 1: The total amount of carbon and nitrogen necessary for each individual per diem must be known; a table also is required giving the amount of C and N in a fixed quantity (¢.g., 1 ounce) of the different foods. The N should be to the C inthe ratio of 1 : 15 ; and for ordinary work, 300 grains of N and 4,500 grains of C may be taken as a standard. Example : Find the amount of bread and milk sufficient for the diet of a man at ordinary work. Let x=amount of bread (in ounces). y= Se ek From a table it is found that— 1 ounce of bread contains 116 grains C and 5'5 grains N. 1 ounce of milk contains 30 grains C and 2°8 grains N. Therefore— a ounces bread contain 116% grains Cand 5'5% grains N. y ounces milk contain 30y grains C and 2°8y grains N ; and total C==4,500.* total N = 300. -. total carbon=116x7+ 30v=4,500...... (i-); total nitrogen = 5°5.r+2°8y=300...... (i1.). From these two equations find the values of x and y. Eliminate y by multiplying each equation by the co- efficient of yin the other equation—that is, by multiplying (1). by.2:8,and ii.) Dyi30..) {huss 32432 + 84y=12,600 .. . «.)» (ill). 165x++84y=9,000.,..... (iv.). Subtract (iv.) from (iii.), and the difference is— 159°8% = 3,600 ; Ships Us 0 159°8 ENERGY, EXERCISE, AND DIET 89 . Substitute this value of x in (ii.), thus : | (5°5 X 22°5) + 2"8y'= 300 j 123°75 + 2°8y = 300; 2°8y = 300 — 123°75=176°25 ¢ . V=63. That is, 22} ounces of bread and 63 ounces of milk will supply the necessary amount of carbon and nitrogen. This problem may also be solved graphically as follows: TIO 307 4,500 ees a: (i) Sb tbs 300) ab 4 aXe. (ii.). Since these equations are of the first degree, their graphs will be straight lines. In G.) if v=o, then y=150 and if y=o, then +=39, the points (0, 150) and (39, 0) will thus be on the graph, and the straight line drawn through these points will be the graph of equation (i.). Similarly for (ii.), if =o, then y=107, and if y=o, then 2=5§4°5. Therefore a straight line drawn through the points (0, 107) and (54°5, 0) will be the graph of equation (ii.). These lines will be found to intersect at the point (22°5, 63) which gives the solution of the problem. Example : How much cane-sugar and dry albumen are required in a mixed dietary to furnish 30 grammes of nitrogen and 350 grammes of carbon ?* Let x=number of grammes of albumen. y=number of grammes of cane-sugar. ————$ re a *-D P.H. Exam (Roy, Coll. Phys: Surg.). 90 AIDS TO THE MATHEMATICS OF HYGIENE The formula for cane-sugar (C,.H»90,,) shows that it contains 144 parts by weight of C in 342 parts, or 144 X 100 oe = 42 per cent. (see example on p. 109). Proteids—e.g., albumen—contain approximately 53 per cent. C and 15 per cent. N, that is: 100 grammes albumen contain 53 grammes C; igs xX .. + grammes albumen contain Too Srammes OF 100 grammes cane-sugar contain 42 grammes C ; 42 -*. ¥ grammes Cane-sugar contain ceed grammes C. 100 ~ Total C must be 350 grammes ; a ae 100 * 100 39°} OF,°5 34 1-42) = 35,0008... ws (i.). 100 grammes albumen contain I5 grammes N ; eld Wea .. ¥ grammes albumen contain 7 grammes N, Total N must be 30 grammes ; OF/1157,= 3,000) 16 fat (ii.). Equation (ii,) shows that += me = 200. Substitute this value of x in (i.), thus : (53 X 200)+ 42y = 35,000 ; 10,600 + 42y = 35,000 ; 427 = 35,000— 10,600 = 24,400 $ .. Y= GON 200 grammes of albumen and 581 grammes of cane- _ sugar are therefore required. Method 2: A ‘standard diet’ is required, showing the total amount of each variety of food which is necessary, and ENERGY, EXERCISE, AND DIET gi also a table, g'ving the percentage composition of different foods. A Report—drawn up by a Committee of the Royal Society, at the request of the Board of Trade—was issued in 1917, which stated that ‘a nation, for the most part engaged in active work,’ requires per man per diem : loo grms. proteids, ‘Ioo grms. fat, and 500 grms. carbohydrates. _ (These figures will be adopted here as ‘the standard diet.’) Example: Find the amount of bread, cheese, and butter sufficient for the diet of a man in active work. Let x=amount of bread (in grammes), Ve Ms Cneese. 21 45 o= . butter 55 From a-table (Parkes and Kenwood) it is found that Proteids, Fats, Carbohydrates, per cent. per cent. per cent. Bread contains 8:0 O'S 50 Cheese ,, 28'0 23°0 I Butter ,, 1°5 83°5 I Theretore— Proteids. Fats. Carbohydrates, ; Se | Obs : x grms. bread contain — en ead grms. 100 100 100 28y 239 ef rms. cheese —_ BS a 100 ~— «100 100.) | BGs 88 zgrms. butter _,, ae EE i ae), 92 AIDS TO THE MATHEMATICS OF HYGIENE But the proteids must be equal to a total of 100 grms., the fats Ioo grms., and the carbohydrates 500 grms. 8x 4 28y 4 U52 .. Proteids= OO. 100" Ico =: 100 O° 83° Fats =f $9 2" Zai00; 100 ‘ies 100 Carbohydrates = oes merge rho 500. 100 I00- f00 Clear of fractions by multiplying each of these three equations by Ioo, thus: CAO yeas 6 —1O,OOO mr eats ks (i.). Orb 2237 1263-52 10,000 ae (i1.). BO V ac. 15 O}000 ute horn car aes (i1l.). There are here three equations, with three unknown quantities, and the first stage in the solution consists in reducing these to two equations, with two unknowns. First eliminate + from (i.) and (ii.) by multiplying (ii.) by 8, and (i.) by 0°5, thus : 4v+ 184y+ 668z=80,000.,.... (iv.). Ava tAV 10:75 5,000. ee (Va) Subtract (v.) from (iv.), and the difference is 170) 4+ 0671252 — 7 5.000 Rte. (vi.). Secondly, eliminate + from (i.) and (iii.), by multiplying (i.) by 50, and (ili.) by 8, thus: ADGA AOGY 2147.5 == 500,000 mary lee (vil.) HOOVG OY + 68 == 400,000 61h erie (viil.) Thus are obtained two equations, containing two un- known quantities only——viz., ENERGY, EXERCISE, AND DIET 93 170y+ 667'252=75,000...... (vi.). 13029 @72— 100,600.40 = (os): Next these two equations with two unknown quantities must be reduced to one equation with one unknown. Eliminate y by multiplying (vi.) by 1,392, and (ix.) by 170, thus : 236,640y + 928,8122= 104,400,000...... (x.). 236,640y + 113,907=17,000,000....... (xi.). Subtract (xi.) from (x.), and the difference is 917,4222 = 87,400,000 ; 87,400,000 | SS 1D 917,422 . Substitute this value of z in (1x.), thus : 1,392y + (67 x 95'2) = 100,000 ; 1,392y = 100,000 — 6,378°4 ; = 03,6216; __ 936216 _ 69 1,392 ‘Substitute the found values of y and 2 in (iii.), thus: 504+ 68+ 95'2= 50,000 ; 504 = 49,836'8 ; _ 49,8363 _ a 50 =997- That is, 997 grms. (35 oz.) of bread, 68 grms. (2°4 oz.) of cheese, and 95°2 grms. (3°3 0z.) of butter will supply the requisite amount of proteids, fats, and carbohydrates. What ts the energy avatlable from ‘the standard diet’ ? When oxidised in the body, the energy developed by I grm. water-free proteid=6'1 foot-tons. I grm. 9 ) fat= 13°33 ” 9 £ grim: 1, » carbohydrate = 4°76 foot-tons. 94 AIDS TO THE MATHEMATICS OF HYGIENE Therefore— 100 grms. proteids yield 100 x 6'1 610 foot-tons 100 grms. fats yield 100 x 13°33 4.333 2 500 grms. carbohydrates yield 500 x 4°76=2,3&0 " Total 4,323 So it is seen that the standard diet for ordinary active work yields rather more than 4,000 foot-tons of potential energy. Of these 4,000 foot-tons, 300 foot-tons, on an average, are expended in external mechanical work (p. 87), and 260 foot-tons on internal bodily work (being utilised by the heart and respiratory organs, etc.). Thus the total for internal and external work = 300+ 260 = 560, which is less than 4 of the total energy available from the food. THE FUEL-VALUE OF FOOD. The amount of heat required to raise 1 litre (“e., 1 kilogram) of water through 1° C. is termed a calorie. The heat value of 1 grm. proteid = 4'I calories. * 4 we ocarbohydrate=4:14 .,; 7 ” ” fat =0°3 ” What ts the heat value of the standard diet for ordinary work ? Ioo grms. proteid=100x 4°I = 410 calories. 100 grms, fat = 100 X 9°3 = 930 4; 500 grms. carbohydrate=500x 4° =2,050 ,, Total 3,390 That is, the standard diet yields approximately 3,400 calories. ENERGY, EXERCISE, AND DIET 95 Example: Find the value of a calorie when the unit of weight is I lb., and that of temperature 1° F. 1o07C.=180° F., 2, 1° C.= 18°F. t kilo=2°2 Ibs: 1 cal rie=1 kilo raised 1°C. *, 1 calorie = 2'2: Ibs. raised 1°8° F. =f ID. raised 22x 16=]7° Ff. Therefore a calorie may also be defined as the amount of heat required to raise 1 lb. of water through 4° F. Example : Calculate the heat-value of 1 lb. of butter. Butter has the following percentage composition (p. gr): Proteids 1°5, Fats 83°5, Carbohydrates 1 per cent. .. Proteids = I°5X4'1= 6°15 calories. Fats = 8395. X9°3=770°55. > 5, Carbohydrates= Ix41= 4’! Total 786°80 That is, too grms. butter yield 786'8 calories ; therefore 1lb. (or 453°6 grms.) of butter yields 99 39 786°8 x 453°6 ma = 3,569 calories. That is, the combustion of 1 lb. of butter would provide sufficient heat to raise 3,569 kilograms of water through 1° C., or 3,569 lbs. of water through 4° F. CHAPEERVIT THE CONSTRUCTION OF A HOSPITAL WARD SUPPOSING it is wished to construct a ward to accom- modate twenty patients, allowing 1,000 cubic feet of space per head—that is to say, a ward whose cubic capacity is 20,000 cubic feet. The variations in its possible dimen- sions are endless—e.g., the following would satisfy the conditions as to cubic capacity : Length. Width. Height. I. 100 feet “1.2, 20 feet fan JO feet 20n 5 Ou aan ets ahAO OG, aes nl Ole Sty Clb en WA a | i ele) 4- 32 » vee 25 59 vee 25 yy etc. etc. etc. But some of these would be useless for the accommoda- tion of twenty patients—e.g., if ten beds were placed on each side of a ward constructed as No. 4, each bed (3 feet 2 =3°2 feet of wall-space, and the enormous height (25 feet) is simply wasted space. Obviously, then, the total amount of cubic space required will not help us much; and it may be remarked at the outset that it is not sufficient for the ward, ¢aken as a whole, to satisfy (as the above examples do) the total gross conditions as to capacity, etc., but that it is neces- 96 wide) would only have THE CONSTRUCTION OF A HOSPITAL WARD 097 sary for each individual patient’s portion of the ward to satisfy certain conditions. What these are will now be inquired into more fully. Number of Patients: | The number of patients to be accommodated is of less importance than the consideration how each patient should be accommodated. The maximum number, how- ever, should not exceed 30. | Conditions as to Height: Anything above 14 or 15 feet should be neglected in calculating the necessary cubic space per head, since organic impurities tend to collect in the lower portions of the atmosphere, and excessive height will not counter- balance this, whereas increased space in other dimensions would dilute these. Width: The minimum width should be 24 feet ; allowing 64 feet for the length of each bed, two beds on opposite sides of the ward would take up 13 feet, leaving 24—13=11 feet for the passage down the middle of the ward between the two rows. Length: No limits within reason. If we have a ward containing twenty patients, and it is wished to enlarge it into one accommodating thirty, it is obvious that the only way it could be done would be by adding to the length ; for, as has been already seen, increase in height should not be allowed to provide for the extra amount of cubic space required, and increase in width would only widen the passage down the middle of the ward between the two rows of beds without providing any further bed accom- modation, 7 98 AIDS TO THE MATHEMATICS OF HYGIENE Floor-Space : From what has been said as to height, it will be seen that dilution of respiratory impurities can only be thoroughly carried out by each patient having a certain amount of floor-space, and the minimum should be Ioo square feet per head, and should not be less than ;4 of the cubic space. Amount of Cubic Space : For a general hospital, each patient should have a minimum of 1,200 cubic feet ; if the air be renewed three times per hour, this would give 3,600 cubic feet of fresh air per head per hour. - Windows : It is generally allowed that there should be 1 square foot of window to every 70 cubic feet of space; they should reach downwards to within 23 or 3 feet of the floor, and upwards to within 1 foot of the ceiling. Where a bed is placed between each window, the space between the windows should be at least 1 foot wider than the width of the bed ; as the beds are usually 3 feet wide, the space between the windows must be at least 4 feet. Beds : Three feet should be the minimum distance between adjoining beds. Let us now see how to apply these data in construct- ing a ward of the usual oblong form to accommodate, é.g., twenty-eight ordinary medical patients. Floor-space must be a minimum of Ioo x 28 = 2,800 square feet. Cubic contents must be a minimum of 1,200~x 28 = 33,600 cubic feet. It will be noticed that this exactly fukfls the condition THE CONSTRUCTION OF A HOSPITAL WARD 99 I that floor-space should be at least ie of the cubic space, ; 6 : since aad =2,800. Knowing the cubic space to be 33,600 cubic feet, and the floor-space 2,800 square feet, it will be seen that a height of = 12 feet will satisfy the r) conditions so far. Having obtained the height and the floor-space, the latter must now be divided up into length and breadth. Taking the minimum permissible width— 8 viz., 24 feet—we are left with ee eas feet for the length. Will this length be sufficient? As there are twenty-eight beds, or fourteen on each side, each bed would thus have se = 8'4 feet of wall-space, which would provide for a bed of 3 feet in width, and a space of 8°4 - 3:0 =5°4 feet between each. As a minimum of only 3 feet between each bed has been fixed, 117 feet wiil amply satisfy the conditions as to length. But it may be re- marked incidentally that although the beds might be put closer together, without overstepping the limit laid down, yet it would not be permissible on that account either to put more beds into the ward, or to shorten its length, otherwise each individual would be deprived of a portion of his floor-space and cubic space. The deficiency in the latter could be overcome by raising the height of the ward to 14 or 15 feet (not beyond this) ; but the deficiency in floor-space could not thus be overcome, except by in- creasing the width of the ward, which, of course, could be done. So the conditions are fulfilled in a ward 117°5 feet long, 24 feet wide, and 12 feet high, which provides a distance of more than 5 feet between each bed, If we choose to make the ward 25 feet wide, the length could 100 AIDS TO THE MATHEMATICS OF HYGIENE then be ~ a = 112 feet, which would provide a wall-space of wat feet for each bed ; this would be well over the minimum limit, and satisfy all conditions laid down, Window-Space : I square foot to every 70 cubic feet of space. There must, therefore, be a minimum of S502 — 480 square feet of window-space. If the windows are placed on each side of the ward, 240 square feet of window-space on each side will be required. If a window be placed between each bed, fourteen beds on each side will require thirteen : : a 240 windows, each having a minimum area of a = 18'4 square 3 feet. If each window commences 3 feet from the floor, and reaches to within 1 foot of the ceiling, since height of ward is 12 feet, height of window must be 8 feet, and ae feet will be the necessary width of window. These are the lowest limits allowable, and they might well be made somewhat wider. To what limit in width may we go? It has been said that there must be 4 feet of wall-space between the windows (or 1 foot more than the width of the bed). If width of bed is 3 feet, the required space between the windows would be left by bringing them to within 6 inches of the bed on each side ; and since space between the beds=5'4 feet, the maximum width of window would be 5°4 feet — (2 x 6 inches) = 4’4 feet. The minimum dimensions of the ward which will satisfy all conditions are, therefore : Length, 117°5 feet; breadth, 24 feet; height, 12 feet ; window-space, 480 square feet. If it be decided to have a window between each bed THE CONSTRUCTION OF A HOSPITAL WARD 101 (z.2., twenty-six windows), the minimum size must be 8 feet high and 2°3 feet wide, and the maximum width 4°4 feet. There are numerous other ways in which the ward could be constructed; moreover, the matter of ventilation has not been touched upon ; but the object in view has been to point out, in as simple a manner as possible, the chief features to be borne in mind, and to suggest a method of ‘ setting to work.’ [NoTeE.—For a fever ward, the allowance for each patient should be : eis ray, Floor-space = 140 square feet ; See Cubic space = 2,000 cubic feet ; € eats Wall-space = 12 linear feet. Since the floor-space should be 140 square feet, the height 2,000 of the ward should be at least ne =14 feet, to obtain the necessary 2,000 feet of cubic space.] When estimating the air-space of a room, some deduc- tion should be made for the space occupied by each person living in the room. The average number of cubic feet occupied by each person =} (weight of person in stones). Example : How much window-space would be required for a room 15 feet long, 12 feet wide, and 12 feet high ?* Cubic space of room=15 X 12x 12 = 2,160 cubic feet. Allowing 1 square foot of window-space to every 70 cubic feet of room-space, 2,160 a Fae 31 square feet of window-space required. * D.P.H. Exam. (Roy. Coll. Phys. Surg.). CHAPTER VIII ALCOHOL AND PROOF-SPIRIT Example : What is proof-spirit? How much alcohol ought it to contain (i.) by volume, (ii.) by weight ?* The relationship of alcohol and proof-spirit by volume may be expressed as follows : Alcohol a dee sos 57 avols; Water ... tis aye Bon eA 3 eee ——. Proof-spirit ... He So milOOwss That is, proof-spirit contains 57 per cent. of alcohol by volume. Or, ASS. 321573 100s fifi te Seb ter 6); Therefore P.d. AX 1°76. Similarly : MES rere So, to find the volume of proof-spirit, multiply the * D.P.H. Exam. (Roy. Cell. Phys, Surg.), 102 ALCOHOL AND PROOF-SPIRIT 103 volume of alcohol by 1°76; and, to find the volume of alcohol, divide the volume of proof-spirit by 1°76. [Since 1°76 1s approximately 13, or 4, an almost identical result would be obtained by the simpler method of multi- plying by { or # respectively. | Overproof.—A spirit ‘z° overproof’ means that 100 vols, of the spirit will dilute with -v vols. of water to form 100+ x vols. of P.S. Therefore, to reduce it to proof, «x vols. of water must be added to 100 vols. of the spirit. Underproof.—‘ 1° underproof’ means that too vols. of the spirit contain only as much alcohol as 1oo— # vols. OF Ds Examples: (i.) A sample of whisky contains 70 per cént. by volume of alcohol. How much overproof is it ? P.S.=AxX 1°76=70X 1'°76=122°7. That is, the sample is 22°7° overproof. (ii.) A sample of whisky contains 43 per cent. by volume of alcohol. How much underproof is it? P.S.=A xX 1'76=43X 1'76=75 ; that is, it contains 75 per cent. of proof-spirit, and is, consequently, 100 - 75 =25° underproof. (iii.) A sample is 30° overproof. What percentage of alcohol does it contain ? Being 30° overproof, it will dilute with water to form 130 vols. of P.S. It therefore contains 74 per cent. of alcohol by volume. Se 1044 AIDS TO THE MATHEMATICS OF HYGIENE (iv.) How much alcohol by volume is contained in a sample of spirit 20° underproof ?* Being 20° underproof, it contains only as much alcohol as 80 vols. of P.S. It therefore contains 46 per cent. of alcohol by volume. (v.) A sample of whisky is found to be 4o per cent. underproof. With how much water has it been fraudu- lently adulterated ? 40° underproof means that it contains as much alcohol as 60 vols. of P.S. Now, under the Sale of Foods and Drugs Act Amendment Act, 1879, whisky may be sold 25° underproof—z.e., containing only 75 per cent. proof- spirit. It may therefore be said that— 75 per cent. P.S.=100 per cent. ‘legally pure’ whisky. .. 60 per cent. P.S.= ee to per cent.‘pure’ whisky. The sample therefore contains 80 per cent. of whisky up to legal standard, and 20 per cent. added water. [NoTE.—It follows from Example (ii.) that the Act of 1879 requires a sample of whisky to contain at least 43 per cent. of alcohol by volume. | (vi.) How much spirit at 10° underproof must be added to a spirit at 30° overproof in order to produce a mixture of 5° overproof ? * D.P.H. Exam. (Roy. Coll. Phys. Surg.), ALCOHOL AND PROOF-SPIRIT 105 Let +=amount of spirit required at 10° underproof, . 9oxr then x gallons contain ioe gallons P.S. Let y=amount required at 30° overproof, ad then y gallons are equivalent to gallons P.S. ee 13°Y — amount of P.S. in mixture. I0o.~ =—100 But mixture required consists of ++y gallons at 5° overproof, so y) and ++ y gallons = gallons P.S. oor. oe: (1+y) “160 100 100 | gor + 130y=105x+ Iosy 25VH= 157; ete 25. ass arg That is, 5 gallons of the weaker spirit must be added to 3 gallons of the stronger spirit to produce the required mixture. This example may also be solved by the use of the ‘rule’ shown on p. 112. rt re ne er Relationship between Volume and Weight. Since specific gravity of alcohol =0'793, ce’ I fluid oz. alcohol weighs 0°793 oz. (say, 0'8 02.) 3 . vol. : weight ;71-: 0°8 ; vol, ~_ 2 " weight coe. e ‘. percentage by vol. =percentage by weight x 1° 26 ; ; Za percentage by weight=percentage by volume xo’8. 106 AIDS TO THE MATHEMATICS OF HYGIENE Therefore, to find percentage of alcohol by weight, multiply the percentage by volume by o'8. And since proof-spirit contains 57 per cent. of alcohol by volume, it contains 57 x o°'8=45'6 per cent. by weight. [NoTE.—This means 45°6 per cent. by weight in 100 volumes, not weight in 100 parts by weight. In roo parts by weight of proof-spirit the alcohol weighs 49°24 parts. | CHAPTER IX CHEMICAL CALCULATIONS Example : Estimate the percentage of SO, in a room after disin- ection with sulphur, 2 Ibs. of sulphur being burnt for every 1,000 cubic feet of air. The equation S,+20,=2S0, shows that 64 parts by weight of S combine with 64 parts of O, to produce 128 parts by weight of SO,; that is, S produces twice its weight of SOs, or 2 lbs. of S produce 4 lbs. of SO,,. 1 cubic foot dry air at 32° F. and 760 mm.=566°85 oe (p. 3). . 1 cubic foot SO, at-32° F. and 760 mm.= =1253°57 grains (p. 4). *, r cubic foot SO, at 60° F. (the ordinary temperature __ 566°85 x 32 14°47 of a room)= 12533°57 a2) \=1185°94 grains (p. 12) ~~ 490-+(60= 32) 2 94 ths =0°166 Th. : 00 that is, I cubic foot pane lb. es 1 Ib: SO.= 5. ~~ oe g cubic feet. *, 4 lbs. ae 6 cubic feet. But 4 lbs. SO, are produced by 2 lbs. of S. *, 2 lbs. S produce 23°6 cubic feet SOx. 107 108 AIDS TO THE MATHEMATICS OF HYGIENE And as 2 lbs. S are burned for every 1,000 cubic feet of air, there are 23°6 cubic feet SO, in 1,000 cubic feet of air, or 2°36 per cent. SO,. Example: What volume of sulphur dioxide (measured at Io° C. and 760 mm.) will be produced by the action of sulphuric acid on 30 grammes of copper ?* The chemical equation is as follows : Cu+2H,SO,=CuSO,+ SO,+2H,0. (63°5) (64) Since the density of SO, <4 = 32 (0553), .*. 32 grammes SO,=11°2 litres (p. 3) ; .. 64 grammes SO,= 22'4 litres. But the chemical equation shows that 64 grammes SO, are evolved by the action of H,SO, on 63°5 grammes Cu. .*. 63°5 grammes Cu produce 22°4 litres SO,; .“. 30 grammes Cu produce Sg 10°58 litres SO, at o° C. and 760 mm. ; and 10°58 litres at o° C. become oe 10°96 litres at 10° C. (p. 6). Ans. : 10°96 litres. Example : How many grains of carbon are contained in 1,000 grains of glucose ? t * D.P.H. Exam. (Roy. Coll. Phys. Surg.). 7 Ibid. CHEMICAL CALCULATIONS 109 Glucose = CgHj20¢. Cy =12x 6=72 Hy= I1X12=12 O, =16X 6=96 I 180 Since there are 72 grains of carbon in 180 grains of glucose, how many will there be in 1,000? BOO": 1,000 212 72.54 1000 X 72 ane — 180 = A400, That is, there are 4oo grains of carbon in I,oco grains of glucose. STANDARD SOLUTIONS. Silver Nitrate Solution for Testing for Chlorine. AgNO, + NaCl=AgCl + NaNO. (Ag = 108, Cl=35°5, AgNO,=170). This equation shows that 35°5 mgrs. Cl combine with 108 mgr. Ag; but 108 mgrs. Ag are contained in 170 mgrs. AgNOs3. .. 35°5 mgrs. Cl require 170 mgrs. AgNO, for chemical combination ; .. I mgr. Cl requires spe 4788 mers. AgNO. And since the solution is generally made of such a strength that I c.c. exactly neutralizes 1 mgr. Cl, each c.c. of solution must contain 4 788 mgrs. AgNOz,, or 1,000 c.c. (t litre) must contain 4°788 grammes AgNO. The standard solution, therefore, is made by dissolving 4°788. grms. AgNO in 1 litre of distilled water, ——- 110 AIDS TO THE MATHEMATICS OF HYGIENE Ammonium Chloride Solution for ‘Nesslerizing.’ The solution is generally made of such a strength that I c.c.=o'01 mgr. of NH3. Since molecular weight of NH,=17, and ” ” NH,Cl=53°5, therefore 17 mgrs. NH3 are contained in 53°5 mers. NH,Cl, 53°5 X OO! | and o'or mgr. NHg, is contained in ar 0'03147 mgr. NH, Cl. Each c.c. therefore must contain 0':03147 mgr. NH,Cl, or 0'03147 grammes per litre. Solution for Estimation of CQ, in Air. Example : In estimating the CO, in air by Pettenkofer’s method, the oxalic acid solution is generally made of such a strength that 1 c.c. represents 0'5 c.c. COs. How is the solution mader The equation H,C,04+ CaQ= CaC,0,+ H,O shows that a molecule of oxalic acid neutralises a molecule of CaO ; but molecular weight of oxalic acid (HyC,04+ 2H,O)=126, and that of CaO= 56, *, 126 parts of oxalic acid neutralise 56 parts of CaO. Again, the equation CO,+ CaO =CaCo, shows that 44 parts of COs, neutralise 56 parts of CaO. Therefore with regard to their neutralising power on CaO 126 grms. oxalic acid= 44 grms. CO,; = 22'4 litres COg (p. 3) ; CHEMICAL CALCULATIONS II! 126 : . a2 Sate grms, oxalic=1 litre COg; .. 5°625 mgrms. oxalic=I c.c. CO,; wee ae 5 y =O°5 c.Cc. COs. And since each c.c. of solution is to represent 0°5 c.c. COs, each c.c. must contain 2°813 mgrms., or 2°813 grms. per litre. ¢ Given Two Solutions of Different Strengths, to make a Third Solution of Intermediate Strength. Let a= percentage strength of weaker solution ; b= es * stronger solution ; c= ie a intermediate solution required. Let «=the required amount of weaker solution; y= os i stronger solution. Then r+y=the amount of intermediate solution formed by the mixture. Since 100 parts of weaker solution contain a parts, x parts contain “ t he rts contain —— parts; Pp fon and since 100 parts of stronger solution contain 4 parts, : arts contain oy arts ; ee Too P ‘ eA ci Oe Ne LOW “. V+y parts contain ae Zon parts. But Ioo parts intermediate solution contain c parts, c(r+y) “. *+y parts contain 106 parts. 112 AIDS TO THE MATHEMATICS OF HYGIENE c(xty)_ ax, by | 100) 2 1001008 ‘PS CEACY = ata oy; x (c—a)=y (6-0), x ._b-¢ y ¢-a Therefore, or > That is, the quantities x and y must be in the proportion of (6-c) : (c—a) respectively, from which the following ‘rule’ is obtained: The amount of weaker solution required ts the difference between the percentage strengths of the stronger and inter- mediate solutions, whilst the amount of stronger solution required ts the difference between those of the intermediate and weaker solutions. - Example : To make a 1 : 40 carbolic acid solution from Acidum Carbolicum Liquefactum B.P. Strength of ac. carb. liq.=90 per cent. I : 40 sol.=2°5 per cent. ae of water =o per cent. amount of ac. carb. liq. =2°5 -o=2'5 parts ; water=90 — 2°5=87'5 parts ; water —87°5_—35. carbeaci mae: ete 9? 93 p) Pd that is, 1 part of carbolic acid and 35 parts of water. CHEMICAL CALCULATIONS 113 To Convert ‘Parts per 100,000’ into ‘Grains per Gallon.’ x parts per 100,000 parts, 4X 70,000 __7x =~ parts per 70,000 parts 100,000 10 P PEE 72 P ; AEs Not : or a grains per 70,000 grains, 7 SNE TA or “> grains per gallon ; that is, + parts per 100,000 parts=7— grains per gallon. So, if ‘grains per gallon’ be multiplied by 10, and divided by 7, the result is ‘ parts per 100,000’ ; and if ‘ parts per 100,000’ be multiplied by 7 and divided by Io, the result is ‘grains per gallon.’ [ NOTE.—1 grain per gallon=1 part per 70,000 parts ; I mgrm. per 100 C.c.=I part per 100,000 parts. ] Example: Express 0°0042 and 0'0053 grain per gallon as parts per million.* 00042 grain per gallon 0042 X I ea oa =0'006 part per 100,000 =0'06 part per 1,009,000, Also, 00063 grain per gallon 0°0063 X 10 aay Sail =0'009 part per 100,000 =0'09 part per 1,000,000. * D. P.M, Exam, (Roy, Coll, Phys. Surg,). 8 114 AIDS TO THE MATHEMATICS OF HYGIENE To Convert ‘Grains per Gallon’ into ‘Parts per 100,000 I galion of water = 70,000 grains. If there are x grains per gallon, there are “¢ grains in 70,000 grains, %X 100,000 __ I On ° ° . 70,0 rae grains In [00,000 grains, yOOO 10x or ays parts per 100,000 parts ; : . 10x that is, x grains per gallon= ca parts per 100,000 parts. NORMAL SOLUTIONS. A normal solution is one which contains the hydrogen equivalent of the active substance in grammes per litre. From the equation HCl+ NaOH =NaCl+H,O itis seen that 36°5 parts of HCl neutralize 4o parts of NaOH ; so that if 36°5 grammes of HCl be added to a litre of water, and 40 grammes of NaOH to another litre, * then equal volumes of these two solutions, when mixed together, will neutralize each other; that is, the solutions are chemically equivalent. These are termed ‘normal’ (*) solutions. From this it follows that a given volume of azy normal acid solution will exactly neutralize the same volume of azy normal alkaline solution. Again, the equation 2HCl + Na,CO,=2NaCl + CO, + H,O shows that 73 parts of HCl neutralize 106 parts of Na.COs,, or 36°5 parts HCl neutralize 53 parts Na,CO, ; that is, 1 molecule of monobasic HCI can neutralize only CHEMICAL CALCULATIONS 15 3 a molecule of bivalent Na,CO,; hence ¥ solution of .. 166 : Na,CO, contains > =53 grammes per litre. Similarly, ¥ solution of acetic acid (H.C.H,O,), which is monobasic (z.¢., contains only 1 molecule of displace- able hydrogen), contains 1+24+3+32=60 grammes oa litre. ¥ solution of tartaric acid (Hy.C,H,O,), which is 7 —_- ~22=75 grammes per litre. And ®& solution of citric acid (H3CsH;O,), which is 3472454112 _ 192 3 3 contains tribasic, contains =64 grammes per _ litre. Example: The acidity of to c.c. of a sample of beer is found to be neutralized by 3 c.c. of decinormal (4,) solution. of NaOH. Express the acidity in terms of glacial acetic acid. From what has been said as to the nature of normal solutions, it is obvious that 3 c.c. of {4 NaOH neutralize 3 c.c. of ,4, acetic ; therefore the acidiny of the sample is Pnalent to 3 c.c. of 44, acetic. But— 7 acetic contains 60 grammes per litre, or 60 mgrs. per c.c .. ayy acetic contains 6 mgrs. per c.c. 42 o c. contain 3X6=18 mers. acetic acid. Therefore the acidity of to c.c. of the beer=18 mers. acetic acid, or 180 mgrs. per 100 C.c. The acidity is sometimes stated as grains per pint, thus ; 116 AIDS TO THE MATHEMATICS OF HYGIENE v 180 mgrs. per 100 C.c. = 180 parts per 100,000 parts 180 X7 10 126 : pe Cua 15°75 grains per pint. = 126 grains per gallon (p. 113). Example : A loaf was found to consist of 33 per cent. of crust, and of 67 per cent. of crumb ; 1c.c. of 1p alkaline solu- tion was required to neutralize the acidity of Io grammes of crust, and 1°75 c.c.;\ alkaline solution were required to neutralize the acidity of Io grammes of crumb. Calcu- late the acidity of the whole loaf, and express the result in terms of acetic acid.* The data given provide only for the percentage, not the actual acidity being estimated. It must be assumed, therefore, that the loaf weighs Ioo grammes, and consists of 33 grms. of crust, and 67 grms. of crumb. Io grms. crust are neutralized by 1 c.c. *, alkaline sol. ; but I c.c. 34, acetic is also neutralized by 1 c.c. +X alk. sol., .. acidity of Io grammes crust =I c.c. , acetic; : " 33. Tf 55 = 2%3=3°3c.c. acetic. Io grms. crumb are neutralized by 1°75 c.c. ¥, alk. sol., .“. acidity of Io grms. crumb=1'75 c.c. 4, acetic; eee S07, as a 67 grms. 4y sa eNO) ORS acetic. Therefore, acidity of (33+67=) 100 grammes loaf= 3°3+11°7=15 cc. 4 acetic. But each c.c. of {4 acetic contains 6 mgrs.; (p. 115). “. 15 c.c. contain 15 X6=go mers. acetic acid; .”. acidity of loo grammes loaf=go mers. acetic acid =O'O9 STMS. 4 459 ee eee * D,P.H, Exam, (Roy. Coll, Phys. Surg.) CHEMICAL CALCULATIONS 117 Acidity of loaf, therefore =o0'o9 per cent. The acidity of bread is sometimes stated in grains per lb., thus :— O'09 per cent. =0'09 grs. per I00 grs. =0'09 X 70 gYrs. per 7000 grs. =6'3 grs. per 7000 grs. =6°3 grs. per lb. CHAPTER X LOGARITHMS IT is impossible in a book of this description, nor is it necessary, to enter into a detailed explanation of logarithms, or to describe their various uses; but if the reader will master the few points here given, he should have no difficulty in making use of the logarithmic (or Registrar-General’s) method of estimating a population. Definition.—A logarithm of a number to a given base 1S the index of the power to which the base must be raised to make it equal to the number. For example, the log of 100 to the base Io (written logy) 100) is 2, since 2 is the index of the power to which the base 10 must be raised to make it equal to the number 100, That is— logy) 100=2, since (10)?= 100. Similarly : log;) 1000 = 3, since (10)?= 1000, and /ogi9 1o=1, since (10)'= Io, and /ogj) I=0, since (10)°=I. In common logs the base is always 10, and may, there- fore, be omitted. Thus, ‘/og Ioo’ is understood to mean *l0g%9 100.’ 118 LOGARITHMS 119 To obtain a working knowledge of logarithms, it is only essential to remember the four fundamental proper- ties which they possess (without burdening the mind as to the proof thereof). These are— (i.) deg (ax 6)=log at log b. (i) Zog ( al =log a—log b. (ili.) log (a)*=n log a, log a (iv.) log R/a = : n The four following examples will illustrate the use of these properties : (i.) Multiplication of numbers can be performed by addition of their logarithms. Example : Multiply 203 by 45. Since log (ax 6) =log a+ log b, os log (203 X 45) =log 203 +log 45. From a table it is found that— log 203 =2°3074960 log 45=1°6532125 By addition 3°9607085 ws log (203 X 45) = 3°9607085. But from the table— 3°9607085 = log 9135 5 . log (203 X 45)=log 91355 *. 203% 45 =9135. (The method of obtaining these logs from the table is explained below.) : 120 AIDS TO THE MATHEMATICS OF HYGIENE (ii.) Division of numbers can be performed by sub- traction of their logs. Example : Divide 203 by 45. ° a Since Jog (5) =log a—log b, . 2o5\e = . log ( 45 )=log 203 — log 45. From the table— log 203 =2°3074960 log 45=1°6532125 By subtraction 0°6542835 ‘ CVS \ Mee agen ee ile ( 45 ) 0°65.42835. But from the table it is found that— 0°6542835 =log 4°51; 203 ws log (72) =log 4°51; (iii.) Involution of numbers, or the process by which the powers of quantities are obtained, can be performed by multiplication of their logs. Example : Find the fourth power of 13; that is, find the value of (13)4. Since log (af=n log a, *. log (13)4=4 log 13. From a table, it is found that— log 13=1°1139434 Multiply by 4 4 “. 4 log 13=4'4557736 os log (13)*=4°4557730. LOGARITHMS 121 But by table — 44557730 = log 28561 $ °. dog (13)$=log 28561 ; oo) (13)*= 28.568. (iv.) Evolution of numbers, or the extraction of roots, can be performed by division of their logs. Example: Find the cuhe root of 3,375. Since log “Ja = ne, — lo = EN From table— log 3375 =3°5282738 ; os PESTS = 11760913; *. log 4/3375 =1'176001 3. But from table— i'l 760913 = log 155 vlog 13375 =log 15; o /3375=15. METHOD OF USING THE TABLES. A logarithm consists of two parts, an integral part before the decimal point, and a decimal part after the point. Thus, dog 1019=3°0081742, of which the figure 3 is the integral part (called the ‘ characteristic’ or ‘index’), and 0081742 the decimal part (called the ‘ mantissa’). In the tables only the decimal part is given; the index must be prefixed according to rule, as explained below. 122 AIDS TO THE MATHEMATICS OF HYGIENE Supposing the logs of the foilowing numbers are required : I°O19, I0°I9, I01°9, 1019, and Io1go. It will be noticed that the digits of these five numbers are all similar, the only difference being the position of the decimal point. Where the digits of the numbers are the same, the mantissa of the logarithm is the same, no matter where the decimal point is placed. It is different with the characteristic, which is always one less than the number of integral figures in the number. Thus, the above five numbers have respectively 1, 2, 3, 4, and 5 integral figures (z.¢., figures in front of the decimal point) ; the characteristics of their logs, therefore, are 0, 1, 2, 3, and 4 respectively, whilst the mantissa is the same in all of them. Therefore— log 1'019=0°0081742 log 10°19=1'0081742 log 101'9=2'0081742 log 1019==3'0081742 log 10190= 4'0081742 Thus, the mantissa is altogether independent of the position of the decimal point, whilst the characteristic is decided solely by that position. | No. ° I 2 3 4 5 6 7 8 9 1014 | 006 0380 0808 1236 1664 2092 2521 2949 3377 3805 4233 15 4660 5088 5516 50944 6372 6799 7227 7655 8082 8510 16 8937 9365 9792-0219 0647 1074 50% 1928 2355 2782 | 27,007 3280 | 3037 4064 4499 4917 5344 577% ea Pot 7051 18 7473 7904 8331 8757 9184 9610 0037 0463 0889 1316 19 | 008 1742 2168 2594 3020 3446 3872 4298 4724 5150 5576 | The above is an extract from a table of logarithms em- bracing all numbers up to five figures, the first four of which will be found in the left-hand column, and the fifth LOGARITHMS 123 at the top. The first line of the table, opposite the number 1014, reads as follows: 0060380, 0060808, — 0061236, etc., the three figures (006) at the commence- ment of the line applying throughout. In the line opposite the number 1016 it will be noticed that some of the figures have a ‘vinculum,’ or line over them, which means that all such figures belong to the next (viz., the 007) group. This line, then, reads as follows : 0068937,.+++-- 0069792, 0070219, 0070647, etc. Simi- larly, the line opposite the number 1o18 reads : 0077478 Dees 0079610, 0080037, etc. To find the Log of a Given Number from the Table. Example : From the table given above, find the log of 101°76. First find the log of 10176, neglecting the decimal point. Opposite 1017 in the first column, and under the figure 6 at the top, will be found 5771 ; prefix the three figures 007, and 0075771 is the mantissa required. What characteristic should there be? The number 101°76 has three integral figures, therefore the characteristic will be 2; hence— log 101°76=2'0075771 (whilst Zog 1017°6 would be 3'0075771). Example : Find the log of 1'0187, Ans. : 0°0080463. To find the Number corresponding to a Given Log. Example : From the given table find the number whose log is 1°0069792. 124 AIDS TO THE MATHEMATICS OF HYGIENE Look in the ‘006 group’ for the figures 9792, which will be found opposite the numbcr ro16, and under the figure 2 at the top. The figures required, therefore, are 10162. Where must the decimal point be placed? Since the characteristic of the given log is 1, the number of integral figures must be two; place the decimal point accordingly, and 10°162 is the number required ; that is— log 10°162 = 1'0069792. Example : % = re < Find the number whose log is 2°0072355. Ans.: 101°68. A table of logs usually extends to numbers of five figures only. Where the numbers contain more than fiv2 frgures, the procedure is as follows: To find the Log of a Given Number which is not Contained in the Tables. Example: Required the log of 1015543. This number lies between 1015500 and 1015600. Take from the table the logs of the first five figures; that is, find the logs of 10155 and 10156, thus: log 10156 = 40067227 log 10155 = 4'0066799 Therefore, /og 1015600=6'0067227 and log 1015500=6'0066799 By subtraction 100 =0'0000428 Thus, an increase of too in the number produces an increase of o*0000428 in the log. But the difference between 1015543 (the number whose log is required) and 1015500 is only 43. What increase will there be in the log corresponding to an increase of 43 in the number ? LOGARITHMS 125 100 : 43 : : 0°0000428 : + ig OE aan k 100 So, éog 1015500=6°0066799 Add 43=0'0000184 Therefore, /og 1015543 =6'0066983 “. 6°0066983 is the log required, Example : Find the log of 101°4736. | Ans. : 2°006353I. To find the Number corresponding to a Given Log which is not exactly contained in the Table. Example : Find the number whose log is 3°0070421. On reference to the table, the mantissa of this log will be seen to lie between 0070219 and co70647, and the numbers opposite these logs are 10163 and 10164 re- spectively. Since the characteristic of the given log is 3, the number required will have four integral figures. Place the decimal point accordingly, thus: log 1016°4 = 3'0070647 log 1016°3 = 3'0070219 Difference, 0°11 =0°0000428 Thus, an increase of 0'0000428 in the log produces an increase of o'1 in the number. But the difference between 3°0070421 (the log whose number is required) and 3°0070219 is only 00000202. What increase in the number will an increase of 0'0000202 produce ? 126 AIDS TO THE MATHEMATICS OF HYGIENE ¢ 00000428 : 0°0000202:: 01: x 0°0000202 X O°! 0°0000428 | Oia Es So, log 1016°3, =3'0070219 add 047 = 0'0000202 . log 1016°347 = 3°0070421 *, 1016°347 is the number required. Example: Find the number whose log is 1'0078925. Ans. : 10°18339. In practice this labour is avoided by the use of the table of ‘ proportional parts,’ in which the necessary multiplication or division is already effected. To find the Log of a Number consisting wholly of a Decimal. Find Jog &. log 3=log 5—log 8; =0°69897 — 0'90309 ; = —0°20412. But £=0'625, and if the log of 625 be sought in a table, the mantissa given there will be found to be 79588. The explanation of this is that in logarithms (‘he mantissa ts always positive, whilst the index may be either positive or negative. Therefore —o'20412 must be expressed in such a way that its mantissa shall be positive. | LOGARITHMS 127 Thus, —0'20412 may be written (-—1+1)-—0'20412 — 1 +(1—0°20412) —1+0°79588 = 1°79588 (the negative sign being placed above the index, to indicate that it refers to that figure alone), ] So, fog 625 =2°79588; log 62°55 =1'79588; log 6°25 =0°79588 ; log 0°625 = 1°79588 ; log 0°0625 = 2°79588 ; It is thus seen that the mantissa is always positive, whether the index is positive or negative. It has been already shown (p. 122) how to find the index in the case of numbers containing one or more integral figures. Where the number consists wholly of a decimal portion, the rule is: Count from the decimal point to the first significant figure ; this gives the index required, which is negative ; thus, Zog 0625 has an index of 1, and the index of Jog 0'00625 is 3, because the first significant figure (viz., 6) is the third from the decimal point. To find the Number corresponding to a Given Log with a Negative Index. Example : ‘ Find the number whose log is 30080889. If the mantissa of this log be looked for in the table on p. 122, the number will be found to be 10188; place the decimal point and sufficient cyphers following it to make the first figure (viz., 1) the third from the decimal point, thus : o’oo10188 ; that is, og 0°0010188 = 3°0080889. CHAPTER XI VITAL STATISTICS ESTIMATION OF POPULATION. THE increase in a population takes place in Geometrical Progression. [NoTE.—(a) Quantities are said to bein arithmetical progres- sion when they increase or decrease by acommon difference ; thus, the numbers 2, 4, 6, 8, ro are in A.P., the common difference being 2. (b) Quantities are said to be in geometrical progression when each is equal to the product of the preceding and some con- stant factor; thus, the numbers 2, 4, 8, 16, 32 arein G.P., the constant factor (or common ratio) being 2. ] Assume that in the year 1901 the population of a town was 10,000, and 13,000 in 1911. There is an increase of 3,000 in the ten years, or an average of 300 annually. But it would be incorrect to say that 10,000 persons become 10,300 by the end of the first year, 10,600 by the end of the second year, and so on, to 13,000 in the tenth year; for the population is increasing year by year, owing to increased number of parents each year, and increased number attaining a marriageable age, and yet this in- creasing number only produces the same stationary annual increase of 300. For instance, if in the first year 10,000 produce an additional 300, making a total of 10,300, the 10,300 commencing the second year (if in- 128 VITAE STATISTICS 129 creasing in the same ratio, which is presunied to be the case) ought to produce more than 300—v2z. 300 X 10,300 10,000 the accompanying table, the population in Example 1 is = 309, and similarly for any other year. In { | Example (in A.P.). || Example 2 (in G.P.). Year. Fopulation. Be Population. rei ae 1gol 10,000 10,000 300 266 1go2 10,300 10,266 300 273 1903 10,600 10,539 300 280 1904 10,900 10,819 300 287 1905 11,200 11,106 * 300 296 1926 11,500 11,402 300 393 1997 11,800 t F705 300 311 1908 12,100 12,016 300 319 1909 12,400 12,335 300 329 I9IO 12,700 12,664 300 2. 336 IQiI I 3,000 13,000 obtained by inserting 9 arithmetical means, and that in Example 2 by inserting 9 geometrical means between 10,000 and 13,000. The column headed ‘ Annual Increase’ shows the difference between the population of any year and that of the preceding one. It will be noticed in Example 2 that the annual addition to the population 5 130 AIDS TO THE MATHEMATICS OF HYGIENE increases with each succeeding year, as would naturally be the case; ¢.g., if 10,000 produce 266, the 10,266 com- 266 x 10266 _ 10000 oy mencing the second year would produce ee This example, therefore, gives the correct estimation of the population, whilst Example 1 shows an incorrect one. It is thus seen that the increase in population takes place _ in Geometrical Progression, : [The reader will doubtless.observe the analogy between these two examples and similar instances of simple and compound interest respectively. If £10,000 be invested at 3 per cent. simple interest, the income will be £300 per annum, or 43,000 for the Io years ; whereas if it be put out at 2°66 per cent., the first year’s interest will only be £266. But if the interest be capitalised each year—that is, if the original sum be invested at 2°66 per cent. compound interest—the amount at the end of Io years will be £13,000.] Methods of Estimating a Population. There are three methods made use of for the purpose of estimating a population. Method 1: This method estimates the population from the average birth-rate of the last ten years. Example : The average birth-rate of a town for the ten years (1907-1916) was 29°2 per 1,000. The actual number of births registered in 1917 was 440. Find the population for 1917. ~ VITAL STATISTICS - 31 There were 29'2 births in 1000 persons, Stet MOCO Or 1 birth in ——— 3 29'2 cs Ss 1000 and 440 births in = 15,066 ; .. estimated population for 1917 =15,066. : ; registered births for the year x 1,000 See en eee ee ee a phate, populate average birth-rate for last Io years Method 2: This method estimates the population from the number of inhabited houses. Example : At the census of 1911 the population of a town was 11,316, and the number of inhabited houses was 1878, 11316 1878 number of inhabited houses (as ascertained from the rate-book) was 2,014. If this number be multiplied by the average number of persons to each house, the pro- duct will give the estimated population for 1917—Viz., (2,014 x 6°012 = )12,108. .. estimated population for 1917 = 12,108. giving =6'012 persons to each house. In1917 the Method 3: This method (known as the Registrar- General's) assumes that the rate of increase remains constant, and is the same as existed during the previous esas period of ten years. Example : Explain the process by which the population of a district is estimated from the last two census returns.* ere od * D.P.H. Exam. (Roy. Coll, Phys. Surg.). 132 AIDS TO THE MATHEMATICS OF HYGIENE Let P=population in any census year, and y=annual increase per unit of population. . Then, by the end of the year, one person becomes I+7, and P persons will become P(1+7). The second year starts with this increased population—viz., P(1+~7), and thus if 1. person. becomes.1I +7, then P(1+7) persons become P(1+7) x (I+7) =P(1+7). Similarly, at the end of the ninth year there will be P(1 +7)° persons, and at the end of the tenth year they will have become P(1 +7)9x (1 +7)=P(1 +7)", or after m years P(1 +7)”. So, if P’= population required, then P'=P(1+7)*. The rate of increase (7) during any intercensal period of ten years can thus be found from this formula, by putting 2z=10, P’=the population at the later census, and P=the population at the earlier census, To con- vert this into a form suitable for solution, take /ogs of each side, thus : | logP!=log[Px (1+ 7)”] =logP+ log (1+7)* =logP+n log (1+7r) ©. 2 log (1+7)=logP'—logP logP'—logP ey + blogP! — logP. zt 10 4 and /og (1 aria) Next, it is assumed that this same rate of increase con- tinues to hold good up to the next census, so that if P”= VITAL STATISTICS s £33 the required population for any year, then, as before, P’=P’ (1+7)%, and JogP"”=logP' +2 log(t+r) ; logP'’—logP , ite) ; logP’-logP | 10 : This formula furnishes the eifeiee practical rule for estimating a population when the figures for the last two censuses are given : (i.) From dog of last census (P’) subtract /og of previous census (P). (Difference = /og of the ¢ex years’ increase.) (ii.) Divide by Io. (Quotient = /og of annual increase.) 3 (i11.) Multiply by the number of years (7) between the last census and the middle of the year sought. (Product = og of the increase for those years.) (iv.) Add Jog of last census (P’). (Sum =/og of the population required (P”). The census is taken at the beginning of April, but all populations should be estimated to the middle _of the y r, or three months later ; ; that is, the ‘central ’ populations must always be found. but it has been shown that log(i+n”= . logP” =logP’ +n. Example : If P represents the population of a town:at the census of 1871, and P’ its population at that of 1881, what will represent its calculated population for Midsummer, 18g0?* : * .D.P.H, Exam. (Roy. Coll..Phys, Surg.).. 134 AIDS TO THE MATHEMATICS OF HYGIENE Here P’=population required for middle of 1890, and 2= 94 years (z.e., from April 1881 to Midsummer 1890) ; log P’ -logP .”. dog population required = og P’ +9}. = Example : The population of a town at the census of 1901 was 9,935, and at that of 1911 was 12,915; find the population for 1917 (2.e., 6} years later). log pop. 1911 =/og 12,915 = 41110948 log pop. 1901 = log 9,935 = 3°9971514 difference = /og of 10 years’ increase = 0°1139434 divide by 1o=/og annual increase =0'0113943 multiply by 64 S 6°25 product =Zog of 63 years’ increase =0'0712140 *add log pop. 1911 =log 12,915 = 4°1110948 .. Cog pop. 1917 = 4'1823088 But 4°1823088 = /og 15,216, ., pop. for 1917 = 15,216. By giving to # (in the formula on p. 133) successive values of 4, 14, 24, etc., the central populations for each year are obtained; or, what is the same thing, the population for any year can be obtained by adding sats 8 Lge log of annual increase) to the /og of the 10 population of the preceding year. * If the required population had been that for 1907—i.¢., 64 years after the former census—then o'o7t2140 (the Jog of 6} years’ increase) would have to be added to /og popula- tion r9or, instead of, as here, to /og population rgrt. VITAL STATISTICS 135 4 Example : Using the census figures of the see aeui example, find the populations for each of the years of the inter- censal period rgo1—1910. In this example ac Nala 3 oe has already been found to be 0011 3943, and if this be added to the /og of any one year’s population, the sum will give the Jog of the next year’s population. For the year 1891! ‘ottay lg only 4 of the dog of annual increase must be added, : corresponding to the three months between the census and the middle of that year. Thus: , or og of annual increase, log census pop. 1901 = 3'9971514 (as seen above) add (4 X 0°0113943) =0'0028486 .*. log central pop. 1901 = 4:0000000 (= lag 10,000) add 0°0113943 .. og pop. 1902 = 4°0113943 (=/og 10,266) add 0'0113943 .”. Jog pop. 1903 = 4°0227886 (=/og 10,539) add 0°0113943 .*. Zog pop. 1904 = 4°0341829 (=/og 10,819) etc. Therefore the central populations for the years 1901- 19IO ar€: 10,000, 10,266, 10,539, 10,819, etc., and will be found tabulated in full on p. 129 (Example 2). Example : Population at 1901 census = 15,126 ; ” IgIt A 18,472. Find the population for 1919. Ans.: 21,783. 136 AIDS TO THE MATHEMATICS OF HYGIENE To find the Mean Annual Population for a Period of Ten Years. Estimate to the middie of the year the populations for each of the ten years, and take their sum ; this gives the total population for the ten years. Divide by 10, and the quotient will be the mean annual population. Using the previous example, and adding up the central populations for the ten years I90I-IQIo inclusive, as set out on p. 129 (Example 2), their sum will be found to be 112,852 ; the : ; 112852 mean annual population will thus be SiG) ae 1285°2. The sum of the populations for ten years may, how- ever, be found without actually estimating the populations for the individual years, as follows : Let P=central population for 1901; then the sum of the successive populations for the ten years 1901-1910 will be (p. 132) : P+ P(r+7)+ P(1+7)?+ P(r t+7)8+.....6+P(1+7)* (1901) (1902) (1903) (1904) (1910) This is a geometrical progression consisting of Io terms, P being the first term, and (1+7) the common ratio. If the sum of these terms be found by the usual algebraical 10 fe riv'-\) method, the total will be oT. 5. x et But P(1+7)!°=central pop. for 1911, and P=centra’ pop. for 1901 : { central pop.\ _ { central pop. ae of 10 for IQII for 1901 _ J. years’ pop. Tees | (1901-1910) rv VITAL STATISTICS 137 Example : According to the census peruse the population of a place for 1901 was 4,647, and for 191I was 5,811. Find the mean annual population for the ten years 1901-1910. (i.) First find the central populations for 1901 and IQIf. The usual formula ‘p. 133) applies, viz. : . central | {re fs 1 Jog pop. 1911 — Jog pop. 1901 as = op. pop. 1901 | ee 10 log central I log 5811 —/og 4647 ee 19goI }= oS ears 10 This works out as follows : log 5811 = 3°7642509 log 4647 = 36671727 difference =0'0970782 (=J/og Io years’ increase) divide by 10 =0'0097078 (=/og annual increase) multiply by + =0'0024269 (=/og ¢ year’s increase) add Jog 4647. =3°6671727 Ber ntrul Pasiode = 3°6695996 (=log 4673) *, certral pop. for 1901 = 4,673. Similarly, { fog central | _ Os | 1 log pop. 1911 — Jog pop. 1901. [Pop- fori) bee 10 IQII log pop. 1911 =log 5811 = 3°7642509 add Jog } year’s increase =0'0024269 (as above) .. dog central pop. 1911 = 3°7666778 (=J/og 5844) .. central pop. for 1911 =5,844. 138 AIDS TO THE MAIHEMATICS OF HYGIENE (ii.) Find the value of 7. log 5811 — 4647 (p.1 but aE, log annual increase log( +7) = 32) 5 =0°0097078 (vide above) ; .. 202(1 +7) =0'0097078. And from a table it is found that 0'0097078 = log 1'0226; o. L+7=1'0226 ; . Y=0'0220. And since central pop. central pop. {i Pop. | { for I9II lg { for I901 t leita) | — years i {cue pop. fie 5844 — 4673 = 51,8143 the Io years 00226. and mean annual pop. for the Io years I9go01-IgIo. =e =H e1814, See If the same Rate of Increase continues, how long will a Population take to Double Itself? P!=P(1+~7)”, and the population will be doubled when =2P; that is, P(r+7)*=2P, .. (1+7)*=2; . 2 log (i+r)=log 2; log 2 tires log (1 I+r) =. a VITAL STATISTICS 539 Applying this to the example on p. 134, log (1+7)=/og annual increase =0'0113943 3 log 2=0°3010300 ; 0°3010300 ee 7 O'0113043 26°4 yearSe BIRTH-RATES AND DEATH-RATES. Birth-rates and death-rates are reckoned as so many births and deaths per 1,000 inhabitants living, at all ages. Example : There were 243 births in a year amongst a population of 9,500 (estimated to the middle of the year) ; find the annual birth-rate for the year. There were 243 births in 9,500 persons, or births per person, 9500 per p pice Wa bec 9500 therefore the annual birth-rate is 25'6; = 25°6 births per 1,000 persons $ registered births x 1 ,000 or, generally, annual birth-rate = population Similarly, the annual death-rate per 1,000 _registered deaths x 1,000 population This rate is knowr as the general,’ ‘recorded,’ are ‘crude’ death-rate. The Zymotic Death-rate is also stated as a rate per 1,000 living, thus : deaths from zymotic diseases x 1,000 ae population 2 140 AIDS TO THE MATHEMATICS OF HYGIENE the zymotic diseases being small-pox, diphtheria, scarla- tina, measles, whooping-cough, diarrhoea, and fever— enteric, typhus, and continued. Note the effect of wrongly estimating the popula- tion. The greater the denominator of a fraction, the less is its value. Therefore, the greater the population registered births x 1,000 population » the less the in the fraction value of such fraction will be, and consequently the less the birth-rate. Therefore, if a population be over- estimated, the birth-rate and death-rate will each be stated too low, and wice versa. The Infantile Mortality Rate is stated as an annual rate of so many deaths under 1 year of age, to 1,000 births registered during the year. Example : There were 500 births registered during the year, and 70 deaths amongst infants under 1 year of age; find the infantile mortality rate. There were 70 deaths per 500 births, or /°. deaths per birth, 500 0 X 1000 am ies = 140 deaths per 1,000 births ; therefore I.M.R.=140; : eee or, generally, I.M.R. __ deaths under 1 year x 1,000 registered births To find the mean annual birth-rate for a period of 10 years. Estimate the population for each of the Io years, add them together, and divide by 10. The result is the VITAL STATISTICS 141 ‘mean annual population’ (or the method shown on p. 136 may be used). Add together the number of births during the Io years, and divide by I0; this gives the mean number of births per annum. Then, as in the case of the annual birth- -rate, mean annual birth-rate oe mean births x 1,000 | the period of Io years f~ mean annual population’ Similarly, mean annual death-rate for mean deaths x 1,000 period of Io years ~ mean annual population’ Annual Rates for Short Periods. A ‘weekly death-rate’ of 21°8 per 1,000 does not mean that 21°8 deaths per 1,000 inhabitants occurred during that week ; but that if the deaths that week continue at the same rate throughout the year, they will produce an annual death-rate of 21°8 per 1,000. A year consists of 365 days, 5 hours, 48 minutes, 57 seconds. Z.é., 1 year = 365°24226 days, = 52°17747 weeks; a quarter =from go to 92 days; a month=from 26 to 31 days. The following general formula may be used for ascer- taining annual rates from (@) weekly, (6) monthly, and (c) quarterly returns. (az) Weekly : Let 6=births in a week ; # =number of days in a week (viz, 7). 142 AIDS TO THE MATHEMATICS OF HYGIENE Then 4 births in x days= < births per diem, bx 365°24226 x and births per annum, 65° \ and We 35 21220 7008 births per 1,000 per annum. x X population In the case of monthly and quarterly returns, the - above formula will apply, with the following alterations : (4) Monthly : Let 6=number of births in the month, | «=number of days in the month (28 to 31). (c) Quarterly : Let 6=number of births in the quarter, x=number of days in the quarter (90 to 92). And similarly for death-rates. Example : In a population of 101,786 there were 213 deaths in 5 weeks; what was the death-rate per 1,000 per annum ?* 213 deaths in 5 weeks = 42'6 deaths per week ; 42°6 X 365°24226 x 1000 ( 42°6 X'52°17747 x —_ Ol ee 7X 101786 s 101786 =21'8 per 1,000, Yo find the death-rate of a combined district, where the death-rates of the individual districts are known. Let A=population in one district, and x=its death-rate per 1,000, Sak ara 1000 then =total deaths in A. — oD. He bexam, (hoy, Coll Phys.surg,), VITAL STATISTICS 143 Let B= population in second district, and y=its death-rate per 1,000, yx B ; then ——— =total deaths in B 3 1000 ef ils Lire eeu onl = 24 — total deaths in combined 1000 | 1000 district (A+ Le Jat age ge : oes and OCR) = death-rate per unit in A+ B, and ——-- = death-rate per 1,000 in A+B. A+B This expression furnishes an examp'e of what is known as a weighted average, where cach ‘quan ity’ is multip ied by its ‘weight’ (ze. the number of persons or things connected with it), the sum of the products formng the numera or, .nd the sum of the weights the denominator. In this example the ‘quan ities’ are the death-ra es. and the ‘w.igh:s’ the popula ions. Example: One district of atown has a population of 150,000, with a death rate of 21 per 1,000 per annum, and another dis- trict of the same town has a population of 20,000, with a death-rate of only 15. Show hy calculation what is the death-rate of the combined districts.* Ax+ By _ (150,000 x 21) +(20,000 x 15) A+B 150,000 + 20,000 =20°3 per 1,000. That this is the correct death-rate for the combined district may be readily proved; for a death-rate of 21 per 1,000 in one district means 3,150 deaths in the 150,000; and one of 15 per 1,000 in the second district means 300 deaths in the 20,000, or a total of 3,150 + 300 deaths in the 150,000+ 20,000 inhabitants—that is, 3,450 deaths in 170,000, or 20°3 per 1,000. * D.P.H. Exam. (Roy. Coll. Phys, Surg.), 144 AIDS TO THE MATHEMATICS OF HYGIENE It would have been an error to assume that the death- rate of the combined district was the average of the rae eee . I death-rates of the individual districts—viz., a= 18. This is only true when the populations of the Se octiiell districts are equal, for then A=B, stir the expression aed “A+B aie deatherates © 21 and I5 per 1,ooo are simply averages, and ‘when an average is deduced from two or more averages—that is, when an average of averages is taken—there must be the same number of numerical units in each’ (Parkes), If the populations of A and B had been equal—z.e., had contained the same number of units—then the average of 21 and 15—viz., 18—would be the correct death-rate for the combined district, but not otherwise. The error arising from ‘averaging an average’ is well shown in the following : If a man walk and run alternate miles at the respective rates of 4 and 12 miles an hour, his average travelling is 6 miles an hour, not 8 (the mean of 4 and 12, which numbers are only ‘averages’ themselves). For he walks the first mile in 15 minutes and runs the se-ond mile in 5 minutes; he therefore covers 2 miles in 20 minutes. That is, his rate of travelling averages 6 miles an hour, If the combined dis rict consists of three parts, C being the population of the third part, and 2 its death-:ate per T,000, VITAL STATISTICS 145 Ar+By+Cz A+B+C Ezample : One part of a town has a population of 6,000, and a death-rate of 19; a second part has a population of 12,000, and a death-rate of 21 ; and the remaining part has a population of 9,000, and a death-rate of 25. What is the death-rate of the whole town ?* Md ae This may be calculated from the formula jt t given, where y then = death-rate of combined district. é A= 6,000, +=I19, B=12,000, y=21, C= 9,000, 2=25; or independently of the formula, as follows A has 19 deaths per 1,000, 19 X 6000 ; Hig ibe aoraalgiane, 14 deaths in the 6,000 1000 B has 21 deaths per 1,000, , 21 X 12000 : OF == 252 i the-12,006'3 1000 C has 25 deaths per 1,000, 23. 208 =225 in the 9,000 3 that is, (114+252+225) deaths in a total population of (6,000 + 12,000 + 9,000), or 591 deaths in 27,000 ; that is, a 248 per 1,000. 27000 The same formula may be used when the population and death-rate of one part of the district and the whole district are given, and it is required to find the death-rate of the other part. * D,P,H. Exam, (Roy. Ooll, Phys, Surg.). 10 146 AIDS TO THE MATHEMATICS OF HYGIENE Example : ‘The population of a combined district is 14,000, and its death-rate 13 per 1,000; the population of one part is 6,000, and its death-rate Io per 1,000; find the death- rate of the other part. x=10, A=6,000, A+ B= 14,000 3 .. B=8,000. Find y. Ar+By _ : ary iM ee ©. Ar+ By=13(A+B) 13,A4+B)—-A I3 X 14000 6000 xX IO y=*5 Aa Foes = 15°25 Corrected Death-Rates. GUw™ ~ Example : ‘ What do you understand by ‘ ae ‘standard,’ and ‘corrected’ death-rates? How are they obtained ?* In comparing the death-rates of two or more towns, the age and sex-distribution of their respective popula- tions must be considered, for the following reasons: Bch saccs _ Under five and over fifty-five years of age, the death- | _ rate is higher than the combined rate for all ages, whilst | _ between five and fifty-five it is lower. An undue proportion of infants or old people will thus raise the death-rate, independently of all other considera- tions. The following example will tend to emphasise this point: A and B are two towns, having the same numerical population, viz., 100,000, each of which may be divided * D,P.H, Exam, (Roy. Coll. Phys. Surg.), VITAL STATISTICS 147 into two groups— a) those under 5 years of age, and (6) those over 5. A Death-Rate| Popula- Death-Rate per 1,000 tion || per 1,000 ( ) Umder 5 | - 2475 10,000 25 (6) Over § 14°5 90,000 15 for whole Death-rate town The table shows that A has the lower ceath-rate in each group ; it might be assumed, therefore, that it would also have a lower combined death-rate than B. But in A 10 per cent. of the population are under 5, whilst in B only 3 per cent. are under 5; and if the general death- rate for each town be calculated (as shown on p- 139), it will be found that A’s death rate=15°5, whilst B’s =15°3. It is evident that the higher death-rate in A is due solely to the larger proportion of young children contained therein. Again, at nearly every age-group the death-rate of females is lower than that of males; an excess of females, therefore, in .a population will lower the death- rate. Hence, some correction must be made in the death-rate of a town for any such disproportionate distri- bution of age and sex, before its rate is comparable with 148 AIDS TO THE MATHEMATICS OF HYGIENE that of another town. The method of making this correction is as follows: A Population, divided | Death-rate ‘Expected’ for Age and Sex (Z. and W.) Number of Age- (1go1 Census). (1891-1900. ) Deaths. period. M. £, M., F. M, F, O- 5 5,622 yA ict eae 2°3) | 352"50s01tO 5-Io 55345 5,47! 4°3 4°4 | 22°9 | 24'0 10-15 50134 | 5,194 | 24 | 2° | 12°3 | 13°5 15-20 4,961 5,049 50 S74 Lois 1 Oo 20-25 ec etc etc. etc CLC etc 25-35 35-45 45-55 55-65 65-75 75-85 | | Petre Totals | 48,761 | 51,239 | 837. | 909 ‘ a ay: Be a | 100,000 | 1,746 “ — Consider the case of a town which at the last census showed a population of 100,000 inhabitants. Divide up the population into the twelve age-groups, as in the accompanying table, according to the figures obtained from the last census returns, distinguishing the sexes. Apply to the population at each age-group and each sex the death-rate for that particular age and sex which obtained for England and Wales generally during the last intercensal period of ten years, and calculate the number of deaths—‘expected deaths ’—which each such rate produces, thus: | VITAL STATISTICS - £49 Take the males at ages o-5—viz., 5,622. The E. and W. death-rate for males at that age is 62°7 per 1,000, and 62°7 X 5622". f000. 7 amongst the 5,622. Again, between 5-10 years, a rate eee) ‘TOOO this rate would produce 352°5 deaths of 4°3 per I,000 means = 22'9 deaths amongst the 5,345 male population at that age-group. Proceed similarly for each age-group and each sex, when the total deaths thus calculated will be found to be (say) 1,746. In other words, if the deaths amongst the 100,000 inhabitants had occurred at the same rate as obtained in E. and W. generally, there would have been 1746 X 1000 = 17°46 per I,000. 100000 Ce é 1,746 deaths, or This rate—viz., 17°46 per 1,000—is known as the standard death-rate. {The standard death-rate is thus seen to be merely an hypothetical one, calculated on the - assumption that the deaths in the town occur at the same rate as in E. and W. generally. | Having applied the E. and W. death-rate at. each group to the population of the town, the standard death- rate of the town thus obtained ought to be the same as in E. and W., other things being equal. But this is not the case, the death-rate for E. and W. during the ten years 1891-1900 having been 182; therefore, the age and sex distribution of the town are obviously different from that of E. and W. The standard death-rate must, there- fore, be raised in the proportion of— rate for E. and W. _ 18:2 standard death-rate _ 17°46 myo neo! to make it comparable with that of E. and W. The factor for correction, then, is 1'0423 ; and if the Iso AIDS TO THE MATHEMATICS OF HYGIENE recorded (z.e., the general, or crude) death-rate in any year be multiplied by this factor, the corrected death- rate is obtained ; or, _ f eorrected) recorded tate a eee x factor. If, for example, the recorded death-rate of the town for 1905 was 16°2, then the corrected death-rate for that year would be 16°2 X 1°0423 = 16°9. Age and sex being thus corrected for, any difference in the rates of the two towns compared may be put down to influences appertaining to the place. Unless the industrial character of the town undergoes some important change, it is found that the age and sex distribution, expressed as a ratio to the total population, remain fairly constant during an intercensal period. For this reason, the ‘factor’ is only calculated once every ten years, the last factor holding good until the next | census provides new figures from which a new factor may be formed. A Standard Million. It has been seen that the standard death-rate is obtained by applying the E. and W. death-rate to the population of the town, divided into sex and age- yroups. The tollowing is an alternative method of correcting the death-rate, thereby eliminating the influences of varying age and sex distribution. The census returns for 1911 showed that the total population of England and Wales was 36,070,492, and that the males under 5 numbered 1,896,041. How many males under 5 would there have been, if the total population had been 1,000 000? 1,000,000 tae = 52,504. 1,896,041 x 2. = 9/04 * 36,070,492 VITAL STATISTICS — | 151 Similarly the males between 5-10 were stated to be 1,852,192 ; therefore in a population of 1,000,000 they would have been 1,000,000 36,070,492 By making a similar calculation for each age-group and sex, a table may be constructed, which shows what is termed a standard million. And since the age and sex distribution in this million is exactly proportional to that of England and Wales, the table may be considered to represent a ‘miniature England and Wales’ reduced in scale approximately 4. Next, apply to the population at each age-group a sex of this miliion the respective death-rate, for each group, found to exist in the town under consideration, and thus estimate what would have been the general death- rate in England and Wales had each age-group of its population been subject to the same mortality as that which prevailed in this town. Thus, of the two methods of correcting the death-rate- the former (p. 149) applies the England and Wales death- rate to the population of the town, whilst the latter method applies the death-rate of the town to a miniature population of England and Wales. 1,852,192 X = 51,349. Comparative Mortality Figure. Example: Explain fully what is meant by the term ‘Comparative Mortality Figure’ as used by the Registrar-General in connection with (2) general mortality, (4) mortality from occupation.* (a) General Mortality: The corrected death-rate referred to on p. 150 may be —$_—__= * D.P.H. Exam. (Roy. Coll. Phys. Surg.), 152 AIDS TO THE MATHEMATICS OF HYGIENE stated in another manner: If the death-rate for E. and W. for the same year (1905) be represented by the num- ber 1,000, what number will represent the corrected death-rate of the town ? death-rate | {corrected death- | for E. and W.f ° it rate for town ai aecor rected death-rate x 1,000 ie death-rate for E. and W. * The rate for E. and W. in 1905 was 15'2; SOOO ies Ue O80 Xx £5,000 j hdeckaaee as aegis) This number—1!,112—is the comparative mortality figure, and is the corrected death-rate for the town compared with the death-rate of E. and W., taken as 1,000. It means that the same number of persons which in 1905 produced 1,000 deaths in E. and W. produced 1,112 deaths in this particular town, after correction for differences in age and sex distribution. [NoTE.—In constructing the ‘ factor,’ the death-rate for E. and W. which is made use of is the annual rate for the past ten years. In estimating the C.M.F. for any year, the E. and W. death-rate for that particular year only is used, for purposes of comparison.] (6) Mortality from occupation : The C.M.F. has another use in statistics, in addition to the one already described, since it is a means by which the healthiness. of different occupations may be _ compared. with one another. For this purpose, males between the working age: of 25 and 65 only are considered. Tatham has shown, from the statistics for 1900-1902, that the death-rate of all males between 25 and 65 years of age is 14’08 per 1,000, and 14’08 deaths per 1,000 VITAL STATISTICS: ©. 53 1,000 X T,000 ois Pe persons. The number 71,005, then, forris a standard population, meaning thereby the number of males between 25 and 65 in E. and W., amongst whom 1,co0o deaths occur annually. This standard population is then divided up into four age-groups, which, approximately, are as follows: persons mean I,ooo deaths in Age. Thousands. 25-35 Re aes 26 35-45 < of cases in series series in first second series series. Lbhateis, the error varies inversely as the square root of the number of cases ; and as the value will vary inversely as the error, it follows that the values of two series vary directly as the square roots of the number of cases in the respective series. VITAL STATISTICS 157 Example: In 100 cases, 70 recover, and 30 die. Find the error in the similar, but larger, series of 10,000 cases. ‘ ; nn .70.30 Error in first series =2 ce ar 7 Awe aoe =0'I3 to unity, or 13 per cent. In second series, let x =error, then x : 0°13: : W100 : \/10,000$ 0°13 x NV 100 / 10,000 t= =0'913 to unity, or 1°3 per cent. From the first series it is seen that recoveries may vary between 70+13=83, and 70—13=57; but in the case of the second series, only between 70+1°3=71°3, and 70~ 1°3=68°7. The respective values of the two series are as /t00 : N10,000=1 : 10. Averages and Probable Error. The arithmetical mean is the one most frequently made use of in statistical enquiries, and is obtained as follows: Take the sum of all the numerical values, and divide this by the number of items in the series. Thus, in the series 2, 3, 4, 6, 9, 11, 14, the sum of these figures is 49, and the number of items composing the series is 7. The arithmetical mean, or average, therefore 7 AO aa Wp The approximation ot this average to the truth may be ascertained by finding the probable error, as follows: Gi.) Take the mean of the series. (i.) Find the mean of all the observations above the mean, and subtract the mean from it; the difference is the mean error in excess. 158 AIDS TO THE MATHEMATICS OF HYGIENE (iii.) Find the mean of all the observations below the mean, and subtract this from the mean; the difference is the mean error in deficiency. (iv.) Take the mean of (11.) and (iii.), which gives the mean error. (v.) The probable error is * of the mean error. Thus, in the series given above, the mean is oa he The observations above the mean—that is, above 7—are 9, It, and 14, and their mean is ace =11°33; the mean error in excess is therefore 11°33-7=4°33. The 2+3+4+6 ; SS ee Owes and the mean error in deficiency is 7 — 3°75 =3'25. a 4332 3°25 mean of the observations below 7 is The mean error will be = 3°79, and the prob- able error $ X 3°79 = 2°53. If another series be taken, ¢.¢., 2, 3, 5; 19, 29, 43, Sf, the probable error may be similarly estimated, and will be found to be 14°58. The relative values of these two series will vary in- versely as the squares of the probable errors; that is, Value of Value of Ist faueill ; See eae (14°58)? + (2°53)? gi 33 ae The value of the Ist series is thus 33 times greater than that of the 2nd. The ‘error’ shows how far one is justified in assuming that the same proportions will be repeated in future cases. It will have been gathered that the word ‘error’ is not synonymous with ‘ mistake,’ but expresses the difference between an estimate and an exact measurement. CHAPTER XII LIFE-TABLES Example: How is a life-table constructed, and what are its uses ? * Probability : If an event can happen in @ waysand fail in 4 ways, and all these ways are equally likely to occur, then the prob- ability of its happening is oop and the probability of its _— That is, the probability of an event happening is expressed by a fraction whose numerator is the number representing the number of favourable events, and whose denominator is the total number of possible events, favourable or unfavourable. [Events are mutually exclusive when the supposition that any one takes place is incompatible with the supposition that any other takes place.] When different events are mutually exclusive, the chance that one or other of the different events will occur is the sum of the chances of the separate events. There- fore the chances of the event either happening or failing b ato But the sum of these two fractions=1; and as it 13 certain that the event will either happen or fail, the failing is ° a will be arb ate * D.P.H. Exam. (Roy. Coll. Phys. Surg.). 159 ha eth 160 AIDS TO THE MATHER ATICS OF HYGIENE probability of a certainty is unity. Two probabilities which together make up unity are called ‘ complementary probabilities.’ If f= probability of a person, aged +, surviving a full year, and gx=probability of the person, aged +, dying within the year, then #,+9,=1, and gr=I1-fDx. Example: If there are 1,000 persons aged x, of whom goo survive to the age ++1, what is the probability of any person aged + living a year? Each of the 1,000 persons must either live or die; therefore the total number of possible events = 1,000. Since goo survive the year, the number of favourable events is 900; therefore, goo 9 => sO *— = O0' “TooOMIO rn and g,=1~2=+=o0'1; 900 and since (Brea A . , . number living at end of year * Px number living at beginning of year’ Again, Let P ,*=mean (or central) population at age +—#.e,, _ between the ages x and ++1 ; ad:=the number of deaths which occur at age +; w=the mean (or central) death-rate per unit of population at age x; then, 7 deaths amongst P, persons mean a deaths at per unit of population ; ou eA that is, p lt eo * Dr. Farr’s notation is retained here; actuaries use the symbol ‘Ly.’ —__conenmmeneancen sce iF LIFE-TABLES ° 161 _Now these deaths are assumed to be evenly distributed throughout the year—some of those dying having just reached the age x, and others being nearly ++1 years— a : : so that half of them, —, occur in the earlier part of the Ax. : year, and 4 in the later part; and since P,=pop. at centre of year of age—z.e., after half the deaths have occurred—the population at the beginning of the year may be represented by pees and that at the end of the 2 : year by Pe that is, go no. living atend of year * E pops .- deaths no. living at beginning of year S ie pop. +4 deaths ie a Ae dx es, yard a. a ae But, : 2 pe he 2 a = Es iD Wx eee) ee - 24M, Pot Z 2+ De There are thus three ways of expressing J, —viz., (i. = no. living at end of year : ~~ no. living at beginning of year’ Gis) _ pop. —% deaths | : + pop.+4 deaths’ eee 25 2— Mx om) : P43 +x - Construction of a Short Life-Table by Dr. Farr’s Method. The statistics regarding population, etc., which are available for the construction of an ordinary life-table, are those extracted from the census returns, and issued I 162 AIDS TO THE MATHEMATICS OF HYGIENE by the Registrar-General. These statistical tables give the enumerated populations at each year of age up to five years, and afterwards in groups of five or ten years only. A /#dl life-table gives the expectation of life at each year of age, and for its construction it is necessary either (i.) to determine the population living zz each year of age from the figures for each group of ages, as given in the census returns ; or (ii.) to calculate the probability of living 1 year (Zz) foreach group of ages, and from these grouped probabilities deduce the similar probabilities for each individual age. Either procedure requires a know- ledge of ‘ finite differences,’ or of ‘ graphic interpolation,’ and need not be discussed here. Dr. Farr (in the supplement to the thirty-fifth annual report of the Registrar-General) introduced a ‘short’ method of constructing a life-table, wherein the expecta- tion of life is only calculated for the groups of years which correspond with those of the census returns, no interpola- tion being required. For local life-tables this ‘short’ method gives suffi- ciently accurate results for the greater part of life. Towards the end of the table its results are unreliable, the expectation given being too high. Dr. Hayward, however, has devised one or two modifications in the short method, whereby its results may be made to approximate very remarkably to those of the extended method. For these the reader is referred to his original paper—‘ On Life-Tables: Their Construction and Practi- cal Application ’—published in vol. Ixii. of the Journal of the Royal Statistical Society. They will be briefly re- ferred to later on, after the ordinary construction of the short table, as devised by Dr. Farr, has been described in detail. : LIFE-TABLES 163 Data.—For a life-table constructed on the returns for the ten years 190I-1910 the following data are neces- sary: (i.) The populations at the censuses of 1901 and 1911, divided according to age and sex. ' (i1.) The number of deaths in the ten years 1901-19I0, distinguishing age and sex. These data (for males only) are set out in Tables I. and II. (p. 164). To these may be added— (iii.) The number of births in the ten years, distin- guishing the sexes. (This item is required for deciding on the ‘ radix,’ as explained later.) _ A separate life-table is constructed foreach sex. Only the Males Table will be described here; but it must be understood that every process and calculation shown below must be repeated, substituting the figures and data applying to females for those already used in the case of males. Each age-group must also be treated separately, as an individual population. Preliminary Calculations.—Before commencing the construction of the life-table some preliminary calcula- tions are required, having for their object the furnishing of the requisite data for the formation of the fx column. This column, although frequently not inserted in a life- table, is, in fact, the foundation of the table. These calculations involve the finding of the total populations for the ten years at each age-group—that is, the total number of lives between these ages, subjected to a year’s risk during these ten years. From the total populations thus found the mean annual populations are known. Total Lives at Risk, and Mean Annual Popula- tions.— Each age-group must be considered separately. If, for example, the ages 25-35 be taken, it will be seen from Table I, that between these ages there were 4,647 1644 AIDS TO THE MATHEMATICS OF HYGIENE Table I, Census Returns (Males only). Igol, | 191. 713 | 710 645 587 621 588 2,941 2,800 2,785 | 2,673 | 3,349 4,547 | 5,811 5 | 32308 | 4,529 —55} 25347 | 3,066 | 1,391 | 1,890 i 31,688 $ ! | Total 26,777 — Table 11. Male Deaths im 10 Years (1901-1910). Table 11. Total Male Population for 10 Years. 7,109 6,240 6,334 6,250 5,948 29,398 275102 27,007 29,824 51,814 38,577 26,735 16,158 8,530 2,911 331 The various age-groups read as follows: o-I =age under I; 5-I1o=age 5 and under Io, etc. Table Ve Total Births in the 10 Years 1901-1910, Males Females Total Table lV. LIFE-TABLES 165 living at the 1901 census, and 5,811 living at the 1911 census. It is required to find the total population living between the ages 25-35 during the ten years. This is calculated from the formula shown on p. 137, where this particular example will be found fully worked out, and from which it will be seen that the total population for the ten years at this age-group= 51,814. This calcula- tion must be repeated for each age-group, and the results entered in Table ITI. 2, Column. (Where #,=probability that a person aged + will survive one year.) Take the same age-group again—viz., 25-35. The total population for the ten years is 51,814 (Table III.) ; there- fore the mean annual population is 5181°4; the total number of deaths for the ten years is 352 (Table II.) ; therefore, the mean annual number of deaths is 35°2, the half of which is 17°6 ; and since— ‘pop. —4 deaths 181°4—17°6 5163°8 eee ip we REID 176 751038 pop.+4 deaths’ *°“25 5181°44+17°6 5199 =0°99323- This process must be repeated for each age-group, and the results entered in Table IV. | Alternative Method of jinding pz. If, in the data provided, the mean annual death-rate per 1,000 at each age-group for the Io years be given, instead of the actual number of deaths, the formula 2—-Max Pee. (p. 161) may be used instead of ~,— pop.—% deaths pop.+4 deaths’ out the deaths. Thus, if in-the age-group 25-35 the thus saving the necessity of calculating i { 166 AIDS TO THE MATHEMATICS OF HYGIENE death-rate had been stated as a mean annual ees ee of 5 2 < 1000 ce 5181"4 0'00679 = death-rate per unit, and 7, =0°00679. =) 679 per 1,000 for the 10 years, then : 2 Mx _2-'00679 _ 1'99321 _ Ye 2+iMx 2+0'00679 2'00679 _ 01993233 thus obtaining the same result as by the other method.] . The data are now complete for.the Construction of the Life-Table. Z. Column. The first column to be constructed is the 7, column. Z,,.=the number living who attain the exact age +, whilst Z,+, will be the number who survive to the end of the year, since it is the number who reach the next year of age—viz., ¥+1; and since no. living at end of year no. living at beginning of year’ 2x = la: * D.= = a . *. tpi That is, if the numbet living at the age x be multiplied by the probability of surviving one year from the age +, the product will give the number living at the next year. uence: Lyx pfo=h, Axpy=l, and 4,x f,=/;. It is quite immaterial what number is taken for 4, ‘since the numbers in this column need not be the absolute LIFE-TABLES 16% numbers living, but only relative numbers. The first value in the Zz column—viz., 4, and which is known as the ‘Radix’—is the number of. annual births in the imaginary population, and the succeeding numbers show how many persons out of 4 born alive complete each year of age. It is usual to take 100,000 births, and divide them into males and females in the proportions found to exist during the ten years under consideration. Thug, the total births during the ten years were 18,250 (Table V.), of which 9,286 were males. If there were 9,286 males out of 18,250 births, how many males would there be out of 100,000 births? 15,250 ¢ 100,000: :701286-5 25 . X= 50,884. The Males table will therefore start with 50,884 lives, whilst the companion Females table will start with 100,000 — 50,884= 49,116. The two tables together will thus trace the life-history of 100,000 persons from birth. For this table, therefore, t y= 50,884 5 and since 1,= ly fo, »*. ¢) = 50,884 X 0°83896 = 42,689. Similarly : Ls=1yX py 3 °, ¢, = 38,680 X 0'98858 = 38,238. These numbers, when obtained, are entered in the éx column of the life-table (p. 172). Having found 4, the next step is to find A. The number surviving one year from age 5 is /;x,, and the number out of these (/,x4;) survivors who survive another year is (/, x Js) xf, or 2, x (H5)?. Similarly, 168 AIDS TO THE MATHEMATICS OF HYGIENE the number who survive 5 years from the age 5 will be J. (H2)* ;' that 1s; Lio= ls X (Bs) 5 ZL) = 38,238 x (0°99570)° = 375423- [These examples are best worked out by logarithms, as follows: that is, Lip = 38,238 x (0°99570)° 5 *, log ly =log 38238+ 5 log 099570 = 4°5824952+ (5 X 19981285) = 4°5824952+ T'9906425 SAIS 73 US 7. = log 37423. Similarly, for the age-groups comprising ten years, los X (Dos)! = los 5 *s 235 == 35,365 X (0°99323)? = 33,042. [It will be understood that, in reality, ~ has a different value for each year of age, and in an extended life-table Dn Deere: py are all different. In a short life-table d; (or, strictly, A,-19) must be looked upon as an average or mean value of the five probabilities Z,,..... po, and is supposed to hold good for each of the five years. ] dx Column. (Where d:=the number of deaths which occur at age x—zi.e., between the ages + and ++ 1.) If from the number living at age x at the beginning of the year, the deaths which occur during the year be subtracted, the difference will give the number Hehe at the beginning of the next year ; or, : le —ady= Letty 2.2.) as= Le — batye LIFE-TABLES 166 The d, column is thus formed as follows: Hh=h-4, a,=l,-1, d,=l; — ly ns = los — Tas, etc. For example: Ay = ly - Li; ) *, Ay = 50,884 — 42,689 = 8 195. ree ds=l,—lyo; o. @; = 38,238 — 37,423 =815, etc. P,, Column. (Where P,,=mean population at centre of year.) It has been seen that the number who attain the age x=l,; that is, Z,,=no. reaching the year of age x, whilst by = NO. reaching the end of that year; and since P,,=mean pop. at centre of year of age, therefore P,, is equivalent to 7, ,, ; and since the deaths are assumed to be equally distributed throughout the year, the pop. at the middle of the year will be the arithmetical mean between those beginning and those completing that year of age. Hence Pe ee = For example: pie {, “50, Ege 689 _ Se ee, ly-+ 1, _ 30;285 +38,680 and P,=2 ; = 38,983. For the age-group 5-10, where 5 years are included, the formula is : 170 AIDS TO THE MATHEMATICS OF HYGIENE L,+¢ _ 45 1 419 3 i patisd ares diy . Pye oo ee Azy xs = 189,150. And for the age-group where ro years are included, the formula is ; Po; -95= 254 fos se iter S pga 38885 3308 one = 342,035. Having obtained all these values, as set out in the P, column of the life-table, the Q, column is next to be formed. Q_,, Column. This represents the populations at P_,.+all higher ages. It thus shows the total number of years lived through by the whole population, from the age x to the end of the table, inclusive. That is: ree Dertobeae kes scotch obs Fae stam OOOO G Beginning at the foot of the P.. column, the values of Q, are obtained by successive additions, thus : Pes = 10,365 = Qos 5 add P,,= 52,210; 62,575 = Pys+ Pes = Qs 5 add P,, = 128,165 ; ". 190,740 = Pes + Pys + Pes = Qos, etc. These numbers—viz., 10,365, 62,575, 190,740, etc., are to be entered from below upwards in the Ox column of the life-table. E ~ Column. E, (also written 2) =the complete expectation of life at LIFE.- eee) “2 the age &. oe has been seen “that Q per Boe ieeate number of years which persons at the age 2x will live, from that age and during all succeeding years, to the end of the life-table; and if this total future lifetime be divided equally amongst the survivors at age x—z.e., amongst /, persons—the quotient will give the average after-lifetime of e@ch individual. That is: Bee Cx ee ae a Thus, for the age- eroup 45-55,Q = 667,290, and J, = 29,451; Dy AS 007,290 __ é oe Cg SEIN 22°6 years. “It is thus seen that the expectation of life does not mean the number of years one may reasonably expect to live, but represents the average number of years which persons of a given age in a life-table, taken one with another, actually do live. For this reason, ‘average (or mean) after-lifetime’ is a better name than ‘ expectation of life.’ Summary. _ The various stages in the construction of a life-table by the short method may be summarized as follows : Obtain the necessary data. Deal with each sex separately. Find the total populations for the ten years at each age-group, and from these the mean annual populations. From the total deaths in the Io years find the mean annual number of deaths. Find Z,. for each age-group from the formula __ pop. —+4 deaths a ~ pop. +4 deaths (where ‘ pop.’ = mean annual population for the Io years, and ‘deaths’=mean annual deaths for the Io years). 172 AIDS TO THE MATHEMATICS OF HYGIENE Decide on the radix, or value of 4. Construct the columns of the life-table by means of the formulze, and in the order given below : Z, column: 4,xf,=t,4,3 ad, column: d,=/,—/,4. 3 1s ea : 2 fe E yaipie eis es P, column: P,= Q, column: Q,=Pz+P “+1 E,, column: E= 7 LIFE-TABLE (FOUNDED ON THE MORTALITY OF THE IO YEARS I9QOI-1910.. MALES. x | a | ie7 P, on E | O-I | 8,193 | 50,884 | 46,786 | 2,264,655 | 44°5 I-2 2,446 | 42,689 41,466 2.217, 000 NES LO 2-3 958 | 40,243 | 39,764 | 2,176,403 | 54°0 3-4 605 | 39,285 | 38,9083 | 2,136,639 | 54°4 4-5 442 | 38,680 | 38,459 | 2,097,656 | 54°2 5-10 | 815 | 38,238 | 189,150 | 2,059,197 | 53°8 10-15 | 445 | 37,423 | 186,005 | 1,870,047 | 49°9 15-20 | 699 | 36,978-| 183,142 | 1,684,042 | 45°5 20-25 | 914 | 36,279 | 179,110 | 1,500,900 | 41°3 25-35 | 2,323 | 35,365 | 342,035 | 1,321,790 | 37°3 35-45 | 3,591 | 33,042 | 312,465 979,755 | 29°6 45-55 | 5,099 | 29,451 | 269,015 667,290 | 22°6° 55-65 7,197 24,352 2075535 398,275 16°3 65-75 | 8,675 | 17,155 | 128,165 190,740 | 11°! 75-85 | 6,514 | 8,478 52,210 62,575 | 7°3 85- 1,855 | 1,964 | 10,365 10,365 | 5°2 109 ° ; The reader is reminded that the life-table given above LIFE-TABLES ie is inserted solely for the purpose of demonstrating the fundamental principles of its construction, and he is specially warned against drawing too strict inferences from it, even with the modifications already referred to on p- 162. The most important of Dr. Hayward’s modifica- tions are: (1) The re-estimation of the populations for the first five years of life, the census returns at these ages not being reliable ; (2) a modified method of calculating the latter part of the P, column, whereby the excessive values of P.. towards the end of the table are reduced, and con- sequently also the excessive values of Q, and E.,; (3) a ~ method of calculating fos, thus avoiding the manifest error in the method shown above, which often gives to pos a higher value than to Zgp. é,, or Curtate Expectation of Life. Let the sum of the terms P,+P .,+P)..+ ves.. to the end of the table be represented by the symbol =P , (that Q,=Z=P,); and let.2., +14, +4,,,+ +++. be similarly represented by 2/, |. In an extended life-table, where all the intermediate values of / are given, 2/, 4, will obviously represent the total lives from the beginning of the year of age ++1 to the end of the table, whilst 2P_, will represent the total lives from the middle of the year of age +—or half a year earlier —to the end. The difference, therefore, between SP, and 2d iS half a year’s population ; and this, ee ‘ecimarel amongst the /,. survivors, gives an average of half a year for each life in the population at that age. That is; 174 AIDS TO THE MATHEMATICS OF HYGIENE Pe Bet e =4 year. vast then, gives what is termed the curtate expecta- v tion of life at age x, and is represented by é: ; ‘age Baty == Exe re And since —— = ee = iS li o. €x— Cx =h yeal, ie) or éx=ext3. That is, the curtate expectation of life may be converted HES the complete expectation of life by the addition of 4 year. [This can also be shown as follows : 9 teehee nUSLS Henk tpn lO Ben NAS C TS TE be | Hea p= ere. Ue lett late we , etc. (p. 169) seo Male Seuletatets ) ./ ae 2 lx letitletat.. TLS en 8 @e Cx, or It will thus be seen that the complete expectation of life takes into account the portion of life-time lived in the year in which death occurs—averaged at 4 year—whilst the curtate expectation of life does not. LIFE-TABLES. - 175, Probable Life-time (probable duration of life, vze probable, equation of life) is the age at which any number of children born will be reduced one-half. To ascertain the probable life-time, an extended life- table, showing the number living at each age, is required ; but the method may be demonstrated by means of the short table. On reference to the Z+ column it will be seen thet about the age of 10 years there are approximately 38,000 living, and that these have been reduced to one-half that number about the age 65. Thus the probable life-time at age to is (65—10=) 55 years; that is, it is an exactly even chance whether a child of 10 lives a further 55 years or not—z.e., whether he survives the age 65 or dies before. - Below are appended a few examples which can be solved with the aid of a life-table. Examples : Find the probability of a man aged 35 surviving 20 years—that is, living until he is 55. From the Z, column of the life-table it will be seen that there are 33,042 living at 35, and 24,352 living at 55; therefore the probability of his living from 35 to Se is (p. 159) : 24,352 33,042 within the 20 years is 1 —0°7370=0'263. = 0°7370, whilst the probability that he will die — Find the probability of a boy aged 5 surviving to the age 25—that is, surviving 20 years. The life-table will show that the probability of pe 355305 38,238 is the probability of his dying within that period. surviving is =0'92486, whilst 1 —0'92486=0'07514 176 AIDS TO THE MATHEMATICS OF HYGIENE Find the probability of a father aged 35, and a son aged 5, both surviving 20 years. [The probability that two independent events should both happen, is the product of the separate probabilities of their happening. | > _ It has been seen that the probability of the man sur- viving 20 years is 0'7370, whilst the probability of the boy surviving 20 years 1s 0°92486; therefore the probability of both father and son surviving 20 years 1S 0°7370 x 0'92486 =0'68162. The probability of the father surviving and the son dying during the period is 0'737 x 0'07514=0'05538. The probability of the father dying and the son surviving is 0'263 x 0'92486 =0'24324. The probability of their both dying is 0'263 x0'07514=0°01976; and the probability that both will not die—z.e., that one at least (not specify- ing which) will survive—is 1 —0°01976=0'98024. What is the probability that exactly one (not specifying which) will survive the given period? Probability of father alone surviving =0'05538, and probability of son alone surviving =0'24324 ; and as, by the nature of the question, these events are mutually exclusive (p. 159), the probability that exactly one will survive 1s, therefore, 0°05538+0'24324=0'29862. This, of course, is also the probability that exactly one will die. If, in the above probabilities, the decimal point be moved three places to the right, the chances per 1,000 will be obtained, and may be tabulated as follows : One at least surviving ... «ss 980 in 1,000 Son surviving .,, Ae ee edi LIFE-TABLES Father surviving a Both surviving : One exactly surviving ... Father dying ... es Father dying and son aie Son dying Father surviving and son dying Father and son both dying 737 in 1,000 681 298 263 243 75 £55 20 12 177 CHAPTER XIII MENSURATION, ETC. [ABBREVIATIONS USED ;: d=diameter, y=radius, /=length, b=breadth, Ch=chord, h=height.] Linear : The ratio of the circumference of a circle to tts drameter is constant, and is usually denoted by the symbol ‘7x.’ Its numerical value, however, cannot be stated exactly. ook 22 Approxunately, it is equal to 7? or, more exactly T= 31416. So we have: __ circumference _ circumference diameter 2 |) O79215 12S" S262) 237550) 99° 4°940 | 33°8 |} 12 |10°457/53°6 || 30 | 31°548] 86° 5°302 | 35° 13 | 11°062/55°4 | 35 | 41°827] 95 5°687 | 37°4 114 |11°906]57°2 || 40 | 54°906} 104 | @'O07 | 36:2 15 (12"699150°O Jr 50-1 617082490 6°534 | 410 ||16 |13°635}60°8 || 60 |148'791] 140 6°998 | 42°38 [117 |14°421|62°6 || 70 |233°093] 158 7492 | 44°6 || 18 |15°357/64°4 || 80 | 354°643|176 8'017 | 46°4 [119 |16°346}66'2 || 90 |525'450] 194 8°574 | 48°2 |}20 |17°391|68'0 || 100 |760'000|212 g'165 | 50°0 00D ON ANUAW YN =O Ll For intermediate temperatures, take the mean of the ten- sions at the temperatures above and below; the result, however, will not be absolutely correct, but approximate 186 AIDS TO THE MATHEMATICS OF HYGIENE only—e.g., find the tension ate.60%) 13/608 FF a575?/C. Take the mean of the tensions at 15° C. and 16° C. Thus, 15° C.=12°699 16° C.=13°635 26°334 (pas 13°167 =tension at 15°5° C., or 60° F. 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